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Extra questions for class 10 maths trigonometry.

The trigonometric ratios of the angle A in right triangle ABC are defined as follows: sine ∠A = \\frac{\text{side oppposite to angle A}}{\text{hypotenuse}}) = \\frac{\text{BC}}{\text{AC}}) cosine ∠A = \\frac{\text{side adjacent to angle A}}{\text{hypotenuse}}) = \\frac{\text{AB}}{\text{AC}}) tangent ∠A = \\frac{\text{side oppposite to angle A}}{\text{side adjacent to angle A}}) = \\frac{\text{BC}}{\text{AB}})

The ratios of cosec A, sec A and cot A are the reciprocals of sin A, cos A and tan A respectively. cosecant ∠A = \\frac{\text{1}}{\text{sine of angle A}}) = \\frac{\text{AC}}{\text{BC}}) secant ∠A = \\frac{\text{1}}{\text{cosine of angle A}}) = \\frac{\text{AC}}{\text{AB}}) cotangent ∠A = \\frac{\text{1}}{\text{tangent of angle A}}) = \\frac{\text{AB}}{\text{BC}})

Trigonometric ratios of specific angles

The values of the trigonometric ratios of an angle do not vary with the lengths of the sides of the traiangle, if the angle remains the same.

∴ The value of sin A or cos A never exceeds 1, whereas the value of sec A or cosec A is always greater than or equal to 1

Trigonometric ratios of complementary angles:

The two angles are said to be complementary if their sum equals 90°. The trigonometric ratios of complementary angles are as follows:

  • sin(90° - A) = cos A
  • tan(90° - A) = cot A
  • sec(90° - A) = cosec A
  • cos(90° - A) = sin A
  • cot(90° - A) = tan A
  • cosec(90° - A) = sec A

Trigonometric identities

An equation involving trigonometric ratios of an angle is said to be trigonometric idnetity , if it is true for all values of the angles involved. Some of the key trigonometric identities used in this chapter are as follows:

  • sin 2 A + cos 2 A = 1
  • sec 2 A =1 + tan 2 A for 0° ≤ A ≤ 90°
  • cosec 2 A = 1 + cot 2 A for 0° < A ≤ 90°

Extra Questions for Class 10 Maths - Trigonometry

Consider right triangle ABC, right angled at B. If AC = 17 units and BC = 8 units determine all the trigonometric ratios of angle C.

Given sin A =\\frac{\text{12}}{\text{37}}), find cos A and tan A.

If C and Z are acute angles and that cos C = cos Z prove that ∠C = ∠Z

In ΔABC right angled at B, sin C = \\frac{\text{5}}{\text{13}}). Find : (i) sin A (ii) cos A (iii) cos C

In triangle ABC, right angled at B if sin A = \\frac{\text{1}}{\text{2}}), find the value of 1. sin C cos A - cos C sin A 2. cos A cos C + sin A sin C

In triangle PQR, right angled at Q if PR = 41 units and PQ - QR = 31 find sec 2 R - tan 2 R.

In triangle ABC right angled at B, AB = 12cm and ∠CAB = 60°. Determine the lengths of the other two sides.

In a triangle ABC right angled at B, AB = 5 cm and AC = 10 cm. Determine ∠ BAC and ∠ BCA.

sin (A + B) = 1 and sin (A - B) =\\frac{1}{2}) ; 0° < A + B ≤ 90°; ∠A > ∠B. Find ∠A and ∠B.

Question 10

Find the value of θ in each of the following. θ is an acute angle. (i) 3 sec 2θ = 2√3 (ii) 4 cot 3θ - 4 = 0 (iii) 2 sin 2θ = 1

Question 11

Solve the following: 0 < θ < 90° (i) 2 sin 2 θ = \\frac{3}{2}) (ii) 3 tan 2 θ + 2 = 3 (iii) cos 2 θ - \\frac{1}{4})= \\frac{1}{2})

Question 12

Evaluate the following: (i) cosec 2 45° + tan 2 45° - 3sin 2 90° (ii) cos 60° cos 30° - sin 60° sin 30° (iii) \\frac{\text{tan 60° - tan 30°}}{\text{1 +tan 60° tan 30°}})

Question 13

Find the value of x in each of the following: (i) cosec 3x = \\frac{\text{cot 30° + cot 60°}}{\text{1 + cot 30° cot ⁡60°}}) (ii) cos x = 2 sin 45° cos 45° - sin 30°

Question 14

If θ is an acute angle and \\frac{\text{sinθ + 1 }}{\text{sinθ - 1}}) =\\frac{\text{√3 + 2 }}{\text{√3 - 2 }}), find θ.

Question 15

In an acute angle triangle ABC, sin (A + B - C) = \\frac{\text{1}}{\text{2}}), cot (A - B + C) = 0 and cos (B + C - A) =\\frac{\text{1}}{\text{2}}). What are the values of A, B, and C?

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  • Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

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Class 10 Maths Important Questions Chapter 8 Introduction to Trigonometry: Free PDF Download

CBSE Class 10 Maths Chapter 8 Important Questions revolve around the concept of trigonometric equations at its base. The Class 10 Maths Ch 8 Important Questions by Vedantu come with all the solutions that are drafted in a way, to provide you with a maximum understanding of the subject and figure out the different aspects of trigonometry. The important questions include questions from all the key concepts like basic trigonometry, opposite & adjacent sides in a right-angled triangle, basic trigonometric ratios, and standard values of trigonometric ratios and complementary trigonometric ratios. 

The solutions to these important questions are drafted in an easy-to-understand method. Additionally, the solutions contain step-wise explanations. Download the Class 10 important questions PDF to ensure a deeper understanding of the topic.

Download CBSE Class 10 Maths Important Questions 2023-24 PDF

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Study Important Questions for Class 10 Mathematics Chapter 8 - Introduction to Trigonometry

1. If \[x\cos \theta y\sin \theta =a,\,x\sin \theta +y\cos \theta =b\], Prove that \[{{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\].

\[x\cos \theta y\sin \theta =a\]        …… (1)

\[x\sin \theta +y\cos \theta =b\]        …… (2)

Squaring and adding the equation (1) and (2) on both sides.

\[{{x}^{2}}{{\cos }^{2}}\theta +{{y}^{2}}{{\sin }^{2}}\theta -2xy\cos \theta \sin \theta +{{x}^{2}}{{\sin }^{2}}\theta +{{y}^{2}}{{\cos }^{2}}\theta +2xy\cos \theta \sin \theta ={{a}^{2}}+{{b}^{2}}\]

\[\Rightarrow {{x}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta  \right)+{{y}^{2}}\left( {{\sin }^{2}}\theta +{{\cos }^{2}}\theta  \right)={{a}^{2}}+{{b}^{2}}\]

\[\Rightarrow {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]

\[\therefore {{x}^{2}}+{{y}^{2}}={{a}^{2}}+{{b}^{2}}\]

Hence proved.

2. Prove that \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] Can Never Be Less Than \[2\].

Given: \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] 

We know that, ${{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta $ and ${{\operatorname{cosec}}^{2}}\theta =1+{{\cot }^{2}}\theta $.

\[\Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =1+{{\tan }^{2}}\theta +1+{{\cot }^{2}}\theta \]

\[\Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta =2+{{\tan }^{2}}\theta +{{\cot }^{2}}\theta \]

Therefore, \[{{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta \] can never be less than \[2\].

3. If \[\sin \varphi =\dfrac{1}{2}\], show that \[3\cos \varphi -4{{\cos }^{3}}\varphi =0\].

Given: \[\sin \varphi =\dfrac{1}{2}\]

We know that $\sin 30{}^\circ =\dfrac{1}{2}$.

While comparing the angles of $\sin $, we get  

\[\Rightarrow \varphi ={{30}^{\circ }}\]

Substitute \[\varphi ={{30}^{\circ }}\] to get 

\[3\cos \varphi -4{{\cos }^{3}}\varphi =3\cos \left( 30{}^\circ  \right)-4{{\cos }^{3}}\left( 30{}^\circ  \right)\]

\[\Rightarrow 3\left( \dfrac{\sqrt{3}}{2} \right)-4\left( \dfrac{3\sqrt{3}}{8} \right)\Rightarrow 0\]

Therefore, \[3\cos \varphi -4{{\cos }^{3}}\varphi =0\] .

Hence proved .

4. If \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4\], then Show that \[\tan \varphi =\dfrac{1}{\sqrt{3}}\].

Given: \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4\]

We know that, ${{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi =1$ and \[\tan \theta =\dfrac{\sin \theta }{\cos \theta }\]

Then, \[7{{\sin }^{2}}\varphi +3{{\cos }^{2}}\varphi =4({{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi )\]

\[\Rightarrow 7{{\sin }^{2}}\varphi -4{{\sin }^{2}}\varphi =4{{\cos }^{2}}\varphi -3{{\cos }^{2}}\varphi \]

\[\Rightarrow 3{{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi \]

\[\Rightarrow \dfrac{{{\sin }^{2}}\varphi }{{{\cos }^{2}}\varphi }=\dfrac{1}{3}\]

\[\Rightarrow {{\tan }^{2}}\varphi =\dfrac{1}{3}\]

\[\Rightarrow \tan \varphi =\dfrac{1}{\sqrt{3}}\]

\[\therefore \tan \varphi =\dfrac{1}{\sqrt{3}}\]

5. If \[\cos \varphi +\sin \varphi =\sqrt{2}\cos \varphi \], Prove that \[\cos \varphi -\sin \varphi =\sqrt{2}\sin \varphi \].

Given : \[\cos \varphi +\sin \varphi =\sqrt{2}\cos \varphi \]

Squaring on both sides, we get

\[\Rightarrow {{\left( \cos \varphi +\sin \varphi  \right)}^{2}}=2{{\cos }^{2}}\varphi \]

\[\Rightarrow {{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi +2\cos \varphi \sin \varphi =2{{\cos }^{2}}\varphi \]

$\Rightarrow {{\sin }^{2}}\varphi =2{{\cos }^{2}}\varphi -{{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi $

$\Rightarrow {{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi $

Add ${{\sin }^{2}}\varphi $ on both sides

\[\Rightarrow 2{{\sin }^{2}}\varphi ={{\cos }^{2}}\varphi -2\cos \varphi \sin \varphi +{{\sin }^{2}}\varphi \]

\[\Rightarrow 2{{\sin }^{2}}\varphi ={{\left( \cos \varphi -\sin \varphi  \right)}^{2}}\] 

\[\therefore \cos \varphi -\sin \varphi =\sqrt{2}\sin \varphi \]

6. If \[\tan A+\sin A=m\] and \[\tan A-\sin A=n\], then Show that \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].

Given: 

\[\tan A+\sin A=m\]           …… (1)

\[\tan A-\sin A=n\]            …… (2)

Now to prove \[{{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\].

Take left-hand side

\[{{m}^{2}}-{{n}^{2}}={{\left( \tan A+\sin A \right)}^{2}}-{{\left( \tan A-\sin A \right)}^{2}}\]

\[\Rightarrow {{\tan }^{2}}A+{{\sin }^{2}}A+2\tan A\sin A-{{\tan }^{2}}A-{{\sin }^{2}}A+2\tan A\sin A\]

\[\Rightarrow 4\tan A\sin A\]         

$\therefore {{m}^{2}}-{{n}^{2}}=4\tan A\sin A$      …… (3)

Now take right-hand side

 \[4\sqrt{mn}=4\sqrt{\left( \tan A+\sin A \right)\left( \tan A-\sin A \right)}\]

\[\Rightarrow 4\sqrt{{{\tan }^{2}}A-{{\sin }^{2}}A}\Rightarrow 4\sqrt{\dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}-{{\sin }^{2}}A}\]

\[\Rightarrow 4\sqrt{{{\sin }^{2}}A\left( \dfrac{1}{{{\cos }^{2}}A}-1 \right)}\Rightarrow 4\sin A\sqrt{{{\sec }^{2}}A-1}\]

\[\Rightarrow 4\sin A\sqrt{{{\tan }^{2}}A}\Rightarrow 4\sin A\tan A\]

Hence, $4\sqrt{mn}=4\tan A\sin A$

\[\therefore {{m}^{2}}-{{n}^{2}}=4\sqrt{mn}\]

Hence proved. 

7. If \[\sec A=x+\dfrac{1}{4x}\], then prove that \[\sec A+\tan A=2x\] or \[\dfrac{1}{2x}\].

Given: \[\sec A=x+\dfrac{1}{4x}\]

Squaring on both sides.

\[\Rightarrow {{\sec }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}\]

We know that, \[{{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow 1+{{\tan }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}\]

\[\Rightarrow {{\tan }^{2}}A={{\left( x+\dfrac{1}{4x} \right)}^{2}}-1\]

\[\Rightarrow {{\tan }^{2}}A={{x}^{2}}+\dfrac{1}{16{{x}^{2}}}+\dfrac{1}{2}-1\Rightarrow {{x}^{2}}+\dfrac{1}{16{{x}^{2}}}-\dfrac{1}{2}\Rightarrow {{\left( x-\dfrac{1}{4x} \right)}^{2}}\]

Taking square root on both sides,

\[\Rightarrow \tan A=\pm \left( x-\dfrac{1}{4x} \right)\].

Now, find $\sec A+\tan A$

If $\tan A=x-\dfrac{1}{4x}$ means

$\sec A+\tan A=x+\dfrac{1}{4x}+x-\dfrac{1}{4x}\Rightarrow 2x$

$\therefore \sec A+\tan A=2x$

And if $\tan A=-x+\dfrac{1}{4x}$ means

$\sec A+\tan A=x+\dfrac{1}{4x}-x+\dfrac{1}{4x}\Rightarrow \dfrac{2}{4x}\Rightarrow \dfrac{1}{2x}$

$\therefore \sec A+\tan A=\dfrac{1}{2x}$

8. If \[A,B\]are Acute Angles and \[\sin A=\cos B\], then Find the Value of \[A+B\].

Given: $\sin A=\cos B$

We know that $\sin A=\cos \left( 90{}^\circ -A \right)$

While comparing the values to get

$\cos B=\cos \left( 90{}^\circ -A \right)$

$\Rightarrow B=90{}^\circ -A\Rightarrow A+B=90{}^\circ $

\[\therefore A+B={{90}^{\circ }}\].

9. Evaluate the Following Questions:

a. Solve for \[\phi \], if \[\tan 5\phi =1\].

Given: \[\tan 5\phi =1\]

We know that, ${{\tan }^{-1}}\left( 1 \right)=45{}^\circ $

$5\phi ={{\tan }^{-1}}\left( 1 \right)\Rightarrow 45{}^\circ $

$5\phi =45{}^\circ $

$\phi =\dfrac{45{}^\circ }{5}\Rightarrow 9{}^\circ $

\[\because \phi ={{9}^{\circ }}\]

b. Solve for \[\varphi \] , if \[\dfrac{\sin \varphi }{1+\cos \varphi }+\dfrac{1+\cos \varphi }{\sin \varphi }=4\].

Given: \[\dfrac{\sin \varphi }{1+\cos \varphi }+\dfrac{1+\cos \varphi }{\sin \varphi }=4\]

\[\dfrac{{{\sin }^{2}}\varphi +1+{{\cos }^{2}}\varphi +2\cos \varphi }{\sin \varphi \left( 1+\cos \varphi  \right)}=4\]

\[\dfrac{{{\sin }^{2}}\varphi +{{(1+\cos \varphi )}^{2}}}{\sin \varphi (1+\cos \varphi )}=4\]

\[\dfrac{2+2\cos \varphi }{\sin \varphi \left( 1+\cos \varphi  \right)}=4\]

\[\dfrac{2(1+\cos \varphi )}{\sin \varphi (1+\cos \varphi )}=4\]

\[\dfrac{2}{\sin \varphi }=4\]

\[\sin \varphi =\dfrac{1}{2}\]

We know that, $\sin 30{}^\circ =\dfrac{1}{2}$

\[\sin \varphi =\sin {{30}^{\circ }}\Rightarrow \varphi ={{30}^{\circ }}\]

\[\therefore \varphi ={{30}^{\circ }}\].

10. If \[\dfrac{\cos \alpha }{\cos \beta }=m\] and \[\dfrac{\cos \alpha }{\sin \beta }=n\], show that \[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\].

Given: \[\dfrac{\cos \alpha }{\cos \beta }=m\]         …… (1)

\[\dfrac{\cos \alpha }{\sin \beta }=n\]             …… (2)

Squaring equation (1) and (2). We get,

\[{{m}^{2}}=\dfrac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }\]

\[{{n}^{2}}=\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }\]

Now to prove \[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\] ,

Take left-hand side,  

\[({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta =\left( \dfrac{{{\cos }^{2}}\alpha }{{{\cos }^{2}}\beta }+\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[=\left( \dfrac{{{\cos }^{2}}\alpha {{\sin }^{2}}\beta +{{\cos }^{2}}\alpha {{\cos }^{2}}\beta }{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[={{\cos }^{2}}\alpha \left( \dfrac{1}{{{\cos }^{2}}\beta {{\sin }^{2}}\beta } \right){{\cos }^{2}}\beta \]

\[=\dfrac{{{\cos }^{2}}\alpha }{{{\sin }^{2}}\beta }\]

\[={{n}^{2}}\]

\[\therefore ({{m}^{2}}+{{n}^{2}}){{\cos }^{2}}\beta ={{n}^{2}}\]

11. If \[7\cos ec\varphi -3\cot \varphi =7\], then prove that \[7\cot \varphi -3\cos ec\varphi =3\].

Given: \[7\cos ec\varphi -3\cot \varphi =7\]

Then prove that, \[7\cot \varphi -3\cos ec\varphi =3\]

\[7\cos ec\varphi -3\cot \varphi =7\]

\[49{{\operatorname{cosec}}^{2}}\varphi +9{{\cot }^{2}}\varphi -42\operatorname{cosec}\varphi \cot \varphi =49\]

We know that, ${{\operatorname{cosec}}^{2}}\varphi =1+{{\cot }^{2}}\varphi $ and ${{\cot }^{2}}\varphi ={{\operatorname{cosec}}^{2}}\varphi -1$.

\[49\left( {{\cot }^{2}}\varphi +1 \right)+9\left( {{\operatorname{cosec}}^{2}}\varphi -1 \right)-42\operatorname{cosec}\varphi \cot \varphi =49\]

\[49{{\cot }^{2}}\varphi +49+9{{\operatorname{cosec}}^{2}}\varphi -9-2\left( 3\operatorname{cosec}\varphi \cdot 7\cot \varphi  \right)=49\]

\[{{\left( 7\cot \varphi -3\operatorname{cosec}\varphi  \right)}^{2}}=49-49+9\]

\[{{\left( 7\cot \varphi -3\operatorname{cosec}\varphi  \right)}^{2}}=9\]

Take square root on both sides, we get

\[\therefore 7\cot \varphi -3\operatorname{cosec}\varphi =3\]

12. Prove that \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\].

Given: \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\]

Let us take left-hand side,

\[\begin{align}&2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1 \\ & =2\left( {{\left( {{\sin }^{2}}\varphi  \right)}^{3}}+{{\left( {{\cos }^{2}}\varphi  \right)}^{3}} \right)-3\left( {{\left( {{\sin }^{2}}\varphi  \right)}^{2}}+{{\left( {{\cos}^{2}}\varphi  \right)}^{2}} \right)+1 \\ \end{align}\]

\[=2\left[ {{\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)}^{3}}-3{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi \left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right) \right]-3\left[ {{\left( {{\sin }^{2}}\varphi +{{\cos }^{2}}\varphi  \right)}^{2}}-2{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]+1\]

\[=2\left[ 1-3{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]-3\left[ 1-2{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi  \right]+1\]

\[=2-6{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi -3+6{{\sin }^{2}}\varphi {{\cos }^{2}}\varphi +1\]

Therefore, \[2\left( {{\sin }^{6}}\varphi +{{\cos }^{6}}\varphi  \right)\text{ }3\left( {{\sin }^{4}}\varphi +{{\cos }^{4}}\varphi  \right)+1=0\].

13. If \[\tan \theta =\dfrac{5}{6}\] and \[\theta =\phi ={{90}^{\circ }}\]. What is the value of \[\cot \phi \].

Given: \[\tan \theta =\dfrac{5}{6}\] and \[\theta =\phi ={{90}^{\circ }}\]

We know that, $\tan \theta =\dfrac{1}{\cot \theta }$.

$\cot \phi =\dfrac{1}{\tan \phi }$

$=\dfrac{1}{{5}/{6}\;}$

$=\dfrac{6}{5}$ 

$\therefore \cot \phi =\dfrac{6}{5}$.

14. What is the Value of \[\tan \varphi \] in terms of \[\sin \varphi \] ?

Given: \[\tan \varphi \]

We know that, \[\tan \varphi =\dfrac{\sin \varphi }{\cos \varphi }\] and \[{{\cos }^{2}}\varphi +{{\sin }^{2}}\varphi =1\]

\[\tan \varphi =\dfrac{\sin \varphi }{\cos \varphi }\]

\[\therefore \tan \varphi =\dfrac{\sin \varphi }{\sqrt{1-{{\sin }^{2}}\varphi }}\]

15. If \[\sec \varphi +\tan \varphi =4\], Find the Value of \[\sin \varphi \], \[\cos \varphi \].

Given: \[\sec \varphi +\tan \varphi =4\]

\[\dfrac{1}{\cos \varphi }+\dfrac{\sin \varphi }{\cos \varphi }=4\]

\[\dfrac{1+\sin \varphi }{\cos \varphi }=4\]

$1+\sin \varphi =4\cos \varphi $

${{\left( 1+\sin \varphi  \right)}^{2}}={{\left( 4\cos \varphi  \right)}^{2}}$

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16{{\cos }^{2}}\varphi $

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16\left( 1-{{\sin }^{2}}\varphi  \right)$

$1+2\sin \varphi +{{\sin }^{2}}\varphi =16-16{{\sin }^{2}}\varphi $

$17{{\sin }^{2}}\varphi +2\sin \varphi -15=0$

$17{{\sin }^{2}}\varphi +17\sin \varphi -15\sin \varphi -15=0$

$17\sin \varphi \left( \sin \varphi +1 \right)-15\left( \sin \varphi +1 \right)=0$

$\left( \sin \varphi +1 \right)\left( 17\sin \varphi -15 \right)=0$

If $\sin \varphi +1=0$

Hence, $\sin \varphi =-1$ is not possible.

Then, $17\sin \varphi -15=0$

$\therefore \sin \varphi =\dfrac{15}{17}$

Now find $\cos \varphi $

Substitute the value of $\sin \varphi $

\[1+\dfrac{15}{17}=4\cos \varphi \]

\[\dfrac{32}{17}=4\cos \varphi \]

\[\Rightarrow \cos \varphi =\dfrac{32}{17\left( 4 \right)}\Rightarrow \dfrac{8}{17}\]

\[\therefore \cos \varphi =\dfrac{8}{17}\]

Short Answer Questions (2 Marks)

1. In \[\Delta ABC\] , Right Angled at \[B,AB=24cm,BC=7cm\].

Determine the Following Equations:    

(i) \[\sin A,\cos A\]

Let us draw a right-angled triangle \[ABC\], right angled at \[B\].

A right triangle ABC with AB=24cm and BC=7cm

Using Pythagoras theorem, find $AC$.

\[A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]

\[={{(24)}^{2}}+{{(7)}^{2}}\]

\[=576+49\]

\[\therefore AC=25cm\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{7}{25}\]

\[\therefore \sin A=\dfrac{7}{25}\]

 \[\cos A=\dfrac{AB}{AC}=\dfrac{24}{25}\]

\[\therefore \cos A=\dfrac{24}{25}\]

(ii) \[\sin C,\cos C\]

A right triangle ABC

\[\sin C=\dfrac{AB}{AC}=\dfrac{24}{25}\]       

\[\therefore \sin C=\dfrac{24}{25}\]

\[\cos C=\dfrac{BC}{AC}=\dfrac{7}{25}\]

\[\therefore \cos C=\dfrac{7}{25}\]

2. In Adjoining Figure, Find the Value of \[\tan P-\cot R\].     

A right triangle ABC with AB=24cm and BC=7cm - (2)

Using Pythagoras theorem,

$P{{R}^{2}}=P{{Q}^{2}}+Q{{R}^{2}}$

\[{{(13)}^{2}}={{(12)}^{2}}+Q{{R}^{2}}\]

\[Q{{R}^{2}}=169-144\Rightarrow 25\]

\[\therefore QR=5cm\]

Then find \[\tan P-\cot R\] ,

First find the value of $\tan P$

$\tan P=\dfrac{side\,opp.\,to\,\angle P}{side\,adj.\,to\,\angle P}=\dfrac{QR}{PQ}=\dfrac{5}{12}$

$\therefore \tan P=\dfrac{5}{12}$

Now find the value of $\cot R$

We know that, $\tan R=\dfrac{1}{\cot R}$

For that we need to first find the value of $\tan R$

$\tan R=\dfrac{side\,opp.\,to\,\angle R}{side\,adj.\,to\,\angle R}=\dfrac{PQ}{QR}=\dfrac{12}{5}$

$\therefore \cot R=\dfrac{5}{12}$

Then, 

\[\tan P-\cot R=\dfrac{QR}{PQ}-\dfrac{QR}{PQ}\]

\[\Rightarrow \dfrac{5}{12}-\dfrac{5}{12}\Rightarrow 0\]

\[\therefore \tan P-\cot R=0\].

3. If \[\sin A=\dfrac{3}{4}\], Calculate the Value of \[\cos A\] and \[\tan A\].                                                         

A right triangle ABC with AC=4k and BC=3k

Given that the triangle \[ABC\] in which \[\angle B={{90}^{\circ }}\]

Let us take \[BC=3k\] and \[AC=4k\]

Then using Pythagoras theorem,

\[AB=\sqrt{{{(AC)}^{2}}-{{(BC)}^{2}}}\]

 \[\Rightarrow \sqrt{{{(4k)}^{2}}-{{(3k)}^{2}}}\]

 \[\Rightarrow \sqrt{16k-9k}\]

\[\Rightarrow k\sqrt{7}\]

$\therefore AB=k\sqrt{7}$

Calculate the value of \[\cos A\]  

$\cos A=\dfrac{AB}{AC}=\dfrac{k\sqrt{7}}{4k}=\dfrac{\sqrt{7}}{4}$

$\therefore \cos A=\dfrac{\sqrt{7}}{4}$

And calculate the value of \[\tan A\]

$\tan A=\dfrac{BC}{AB}=\dfrac{3k}{k\sqrt{7}}=\dfrac{3}{\sqrt{7}}$

$\therefore \tan A=\dfrac{3}{\sqrt{7}}$

4. Given \[15\cot A=8\], Find the Values of \[\sin A\] and \[\sec A\].                                                           

Given: \[15\cot A=8\]

Let us assume a triangle \[ABC\] in which \[\angle B={{90}^{\circ }}\]. 

\[15\cot A=8\]

\[\Rightarrow \cot A=\dfrac{8}{15}\]

Since $\cot A=\dfrac{adj}{hyp}=\dfrac{AB}{BC}$.

Let us draw the triangle.

A right triangle ABC with AB=8k and BC=15k

Now, \[AB=8k\] and \[BC=15k\].

Using Pythagoras theorem, find the value of $AC$.

\[AC=\sqrt{{{(AB)}^{2}}+{{(BC)}^{2}}}\]

\[\Rightarrow \sqrt{{{(8k)}^{2}}+{{(15k)}^{2}}}\]

$\Rightarrow \sqrt{64{{k}^{2}}+225{{k}^{2}}}$

$\Rightarrow \sqrt{289{{k}^{2}}}$

\[\Rightarrow 17k\]

$\therefore AC=17k$

Now, find the values of \[\sin A\] and \[\sec A\] .

\[\sin A=\dfrac{BC}{AC}=\dfrac{15k}{17k}=\dfrac{15}{17}\]

$\therefore \sin A=\dfrac{15}{17}$

\[\sec A=\dfrac{AC}{AB}=\dfrac{17k}{8k}=\dfrac{17}{8}\]

 \[\therefore \sec A=\dfrac{17}{8}\]

5. If \[\angle A\] and \[\angle \,B\] are Acute Angles Such That \[\cos A=\cos B\], then show that \[\angle A=\angle \,B\]

A right triangle ABC right angled at C

Given: \[\cos A=\cos B\]

In right triangle \[ABC\],

\[\cos A=\dfrac{side\,adj.\,A}{hyp.}=\dfrac{AC}{AB}\]  …… (1)

And, \[\cos B=\dfrac{side\,adj.\,B}{hyp.}=\dfrac{BC}{AB}\] …… (2)

Then, \[\cos A=\cos B\]

Now, equate equation (1) and (2).

\[\Rightarrow \dfrac{AC}{AB}=\dfrac{BC}{AB}\]

$\Rightarrow AC=BC$

\[\Rightarrow \angle A=\angle B\].

Therefore, Angles opposite to equal sides are equal.

6. State Whether the Following are True or False. Justify Your Answer.           

 The Value of \[\tan A\] is Always Less than \[1\].

False because sides of a right triangle may have any length, so \[\tan A\] may have any value. For example, $\tan A=\dfrac{BC}{AB}=\dfrac{15}{10}=\dfrac{3}{2}=1.5$. 

  \[\sec A=\dfrac{12}{5}\] for Some Value of Angle \[A\] .

True as \[\sec A\] is always greater than \[1\]. For example, $\sec A=\dfrac{hyp.}{side\,adj.\,A}$ . As hypotenuse will be the largest side. So, it is true.

  \[\cos A\] is the Abbreviation Used for the Cosecant of Angle \[A\] .

False as \[\cos A\] is the abbreviation of \[\operatorname{cosineA}\]. Because $\cos A$ means cosine of angle $A$ and $\operatorname{co}\sec A$ means cosecant of angle $A$.

 \[\cot A\] is the Product of \[\cot \]and \[A\].

False as \[\cot A\] is not the product of \[cot\] and \[A\]. \[cot\] without \[A\] doesn’t have meaning. 

 \[\sin \theta =\dfrac{4}{3}\] for Some Angle \[\theta \].

Ans:  

False as \[\sin \theta \] cannot be greater than \[1\]. For example, $\sin \theta =\dfrac{side\,opp.\,\theta }{hyp}$. Since the hypotenuse is the largest side. So, \[\sin \theta \] will be less than $1$.

7. Evaluate the Following Equations: 

i. \[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}\]

Given: \[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}\]

We know that, $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $

\[\dfrac{\sin {{18}^{\circ }}}{\cos {{72}^{\circ }}}=\dfrac{\sin ({{90}^{\circ }}-{{72}^{\circ }})}{\cos {{72}^{\circ }}}=\dfrac{\cos {{72}^{\circ }}}{\cos {{72}^{\circ }}}=1\]                   

$\therefore \dfrac{\sin 18{}^\circ }{\cos 72{}^\circ }=1$

ii. \[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}\]

Given: \[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}\]

We know that, $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $

\[\dfrac{\tan {{26}^{\circ }}}{\cot {{64}^{\circ }}}=\dfrac{\tan ({{90}^{\circ }}-{{64}^{\circ }})}{\cot {{64}^{\circ }}}=\dfrac{\cot {{64}^{\circ }}}{\cot {{64}^{\circ }}}=1\]

$\therefore \dfrac{\tan 26{}^\circ }{\cot 64{}^\circ }=1$

iii. \[\cos {{48}^{\circ }}-\sin {{42}^{\circ }}\]

Given: \[\cos {{48}^{\circ }}-\sin {{42}^{\circ }}\]

We know that, $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $

\[\Rightarrow \cos ({{90}^{\circ }}-{{42}^{\circ }})-\sin {{42}^{\circ }}\]

\[\Rightarrow \sin {{42}^{\circ }}-\sin {{42}^{\circ }}\Rightarrow 0\]

 \[\therefore \cos 48{}^\circ -\sin 42{}^\circ =0\]

iv. \[\operatorname{cosec}31{}^\circ -\sec 59{}^\circ \]

Given: \[\operatorname{cosec}31{}^\circ -\sec 59{}^\circ \]

We know that, $\operatorname{cosec}\left( 90{}^\circ -\theta  \right)=\sec \theta $

\[\Rightarrow \operatorname{cosec}(({{90}^{\circ }}-{{59}^{\circ }})-\sec {{59}^{\circ }}\]

\[\Rightarrow \sec {{59}^{\circ }}-\sec {{59}^{\circ }}\Rightarrow 0\]

\[\therefore \operatorname{cosec}31{}^\circ -\sec 59{}^\circ =0\]

1. Show that the Following Equations:                       

\[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

Given: \[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

We know that, $\tan \left( 90{}^\circ -\theta  \right)=\cot \theta $.

Now let us take left-hand side,

\[\tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \tan ({{90}^{\circ }}-{{42}^{\circ }})\tan ({{90}^{\circ }}-{{67}^{\circ }})\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \cot {{42}^{\circ }}\cot {{67}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow \dfrac{1}{\tan {{42}^{\circ }}}.\dfrac{1}{\tan {{67}^{\circ }}}.\tan {{42}^{\circ }}\tan {{67}^{\circ }}\]

\[\Rightarrow 1\] is equal to R.H.S

\[\therefore \tan {{48}^{\circ }}\tan {{23}^{\circ }}\tan {{42}^{\circ }}\tan {{67}^{\circ }}=1\]

(ii) \[\cos {{38}^{\circ }}\cos {{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0\]

Given: \[\cos {{38}^{\circ }}\cos {{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0\]

Now let us take left-hand side, 

\[\cos {{38}^{\circ }}cos{{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow \cos ({{90}^{\circ }}-{{52}^{\circ }})\cos ({{90}^{\circ }}-{{38}^{\circ }})-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow \sin {{52}^{\circ }}\sin {{38}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}\]

\[\Rightarrow 0\] is equal to R.H.S

$\therefore \cos {{38}^{\circ }}cos{{52}^{\circ }}-\sin {{38}^{\circ }}\sin {{52}^{\circ }}=0$

2. If  \[\tan 2A=\cot (A-{{18}^{\circ }})\] where \[2A\] is an Acute Angle, Find the Value of \[A\]. 

Given: \[\tan 2A=\cot (A-{{18}^{\circ }})\]

We know that, $\cot \left( 90{}^\circ -\theta  \right)=\tan \theta $

\[\Rightarrow \cot ({{90}^{\circ }}-2A)=\cot (A-{{18}^{\circ }})\]

Now equalise the angles,

\[{{90}^{\circ }}-2A=A-{{18}^{\circ }}\]

\[-2A-A=-{{18}^{\circ }}-{{90}^{\circ }}\]

\[-3A=-{{108}^{\circ }}\]

\[A=\dfrac{108{}^\circ }{3}\]

$\therefore A=36{}^\circ $

3. If \[\tan A=\cot B\], then Prove That \[A+B={{90}^{\circ }}\].                                                            

Given: \[\tan A=\cot B\]

\[\cot ({{90}^{\circ }}-A)=\cot B\]

\[{{90}^{\circ }}-A=B\]

\[\Rightarrow A+B={{90}^{\circ }}\]

$\therefore A+B=90{}^\circ $

4. If \[\sec 4A=\operatorname{cosec}(A-{{20}^{\circ }})\], Where \[4A\] is an Acute Angle, Then Find the Value of $A$.

Ans:  

Given: \[\sec 4A=\operatorname{cosec}(A-{{20}^{\circ }})\]

\[\Rightarrow \operatorname{cosec}({{90}^{\circ }}-4A)=\operatorname{cosec}(A-{{20}^{\circ }})\]

\[{{90}^{\circ }}-4A=A-{{20}^{\circ }}\]

\[-4A-A=-{{20}^{\circ }}-{{90}^{\circ }}\]

\[-5A=-{{110}^{\circ }}\]

\[A=\dfrac{{{110}^{\circ }}}{5}\]

\[\therefore A={{22}^{\circ }}\]

5. If \[A,B\] and \[C\] are Interior Angles of a \[\Delta ABC\], then Show That \[\sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}\].

Given: \[A,B\] and \[C\] are interior angles of a \[\Delta ABC\].

We know that, \[A+B+C={{180}^{\circ }}\].

Let us consider, 

\[\dfrac{A+B+C}{2}={{90}^{\circ }}\]

\[\Rightarrow \dfrac{B+C}{2}={{90}^{\circ }}-\dfrac{A}{2}\]

Multiply $\sin $ on both sides,

\[\sin \left( \dfrac{B+C}{2} \right)=\sin \left( {{90}^{\circ }}-\dfrac{A}{2} \right)\]

\[\therefore \sin \left( \dfrac{B+C}{2} \right)=\cos \dfrac{A}{2}\]

6. Express \[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}\] in terms of trigonometric ratios of angles between  \[{{0}^{\circ }}\] and \[{{45}^{\circ }}\].                                                   

Given : \[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}\].

We know that, $\sin \left( 90{}^\circ -\theta  \right)=\cos \theta $ and $\cos \left( 90{}^\circ -\theta  \right)=\sin \theta $.

\[\sin {{67}^{\circ }}+\cos {{75}^{\circ }}=\sin ({{90}^{\circ }}-{{23}^{\circ }})+\cos ({{90}^{\circ }}-{{15}^{\circ }})\]

\[=\cos {{23}^{\circ }}+\sin {{15}^{\circ }}\]

\[\therefore \cos {{23}^{\circ }}+\sin {{15}^{\circ }}\] is the required value.

7. Express the Trigonometric Ratios \[\sin A,\sec A\] and \[\tan A\] in Terms of \[\cot A\]. 

Find the value of \[\sin A\] in terms of \[\cot A\].

By using identity \[{{\operatorname{cosec}}^{2}}A-{{\cot }^{2}}A=1\].

Then, use $\operatorname{cosec}A=\dfrac{1}{\sin A}$

\[\Rightarrow {{\operatorname{cosec}}^{2}}A=1+{{\cot }^{2}}A\]

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}A}=1+{{\cot }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{1}{1+{{\cot }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

\[\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}\]

Now find the value for \[\sec A\] in terms of \[\cot A\].

Using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\sec }^{2}}A=1+{{\tan }^{2}}A\]

\[\Rightarrow {{\sec }^{2}}A=1+\dfrac{1}{{{\cot }^{2}}A}\]

\[\Rightarrow {{\sec }^{2}}A=\dfrac{{{\cot }^{2}}A+1}{{{\cot }^{2}}A}\]

\[\Rightarrow \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\] 

\[\therefore \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

And find the value for \[\tan A\] in terms of \[\cot A\]

By trigonometric ratio property, $\tan A=\dfrac{1}{\cot A}$

Hence, \[\tan A=\dfrac{1}{\cot A}\]

Therefore, \[\sin A,\sec A\] and \[\tan A\] are founded in terms of \[\cot A\].  

8. Write the Other Trigonometric Ratios of $A$ in Terms of \[\sec A\] .                        

Find the value of \[\sin A\] in terms of \[\sec A\]

By using identity,\[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\]

\[\Rightarrow {{\sin }^{2}}A=1-\dfrac{1}{{{\sec }^{2}}A}\]

\[\Rightarrow {{\sin }^{2}}A=\dfrac{{{\sec }^{2}}A-1}{{{\sec }^{2}}A}\]

\[\Rightarrow \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

\[\therefore \sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\]

Now find the value for \[\cos A\] in terms of \[\sec A\],

By trigonometric ratio property, $\cos A=\dfrac{1}{\sec A}$

$\therefore \cos A=\dfrac{1}{\sec A}$

Find the value for $\tan A$ in terms of \[\sec A\],

By using identity, \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\tan }^{2}}A={{\sec }^{2}}A-1\]

\[\Rightarrow \tan A=\sqrt{{{\sec }^{2}}A-1}\]

\[\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}\]

Find the value for \[\operatorname{cosec}A\] in terms of \[\sec A\]

By trigonometric ratio property, \[\operatorname{cosec}A=\dfrac{1}{\sin A}\]

\[\Rightarrow \operatorname{cosec}A=\dfrac{1}{\sin A}\]

Substitute the value of \[\sin A=\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}\].

\[\Rightarrow \operatorname{cosec}A=\dfrac{1}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}\]

\[\Rightarrow \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

\[\therefore \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

Finally, find the value for \[\cot A\] in terms of \[\sec A\]

By trigonometric ratio property, \[\cot A=\dfrac{1}{\tan A}\]

$\Rightarrow \cot A=\dfrac{1}{\tan A}$

Substitute the value of \[\tan A=\sqrt{{{\sec }^{2}}A-1}\]

\[\Rightarrow \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$.

9. Evaluate the Following Equations:                                    

(i) \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

Given: \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and  \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]                      

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$         

(ii) \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

Given: \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

We know that, $\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B$

\[\Rightarrow \sin \left( 25{}^\circ +65{}^\circ  \right)\]

\[\Rightarrow \sin 90{}^\circ \]                    

\[\Rightarrow 1\]

$\therefore \sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}=1$

10. Show that Any Positive Odd Integer Is of the Form \[\mathbf{6q}\text{ }+\text{ }\mathbf{1}\], or \[\mathbf{6q}\text{ }+\text{ }\mathbf{3},\]or \[\mathbf{6q}\text{ }+\text{ }\mathbf{5}\], where \[q\] is some integer.                                                      

Let \[a\] be any positive integer and \[b=\text{ }6\]. 

Then, by Euclid’s algorithm,

\[a=6q+r\] for some integer \[q\ge 0\], and \[r=0,1,2,3,4,5\] because \[0\le r<\text{ }6\].

Therefore, \[a=6q\] or \[6q+\text{ }1\] or \[6q+\text{ }2\]or \[6q\text{ }+3\]or \[6q+\text{ }4\]or \[6q+\text{ }5\]

Also, \[6q+1=2\times 3q+1=2{{k}_{1}}+1,\] where \[{{k}_{1}}\] is a positive integer

\[6q+3=(6q+2)+1=2(3q+1)+1=2{{k}_{2}}+\text{ }1,\]Where \[{{k}_{2}}\]is an integer

\[6q+5=(6q+4)+1=2(3q+2)+1=2{{k}_{3}}+1\], where \[{{k}_{3}}\] is an integer

Clearly, \[6q+1,6q+3,6q+5\] are of the form \[2k+\text{ }1,\]where \[k\]an integer is.

Therefore, \[6q+1,6q+3,6q+5\] are not exactly divisible by \[2\].

Hence, these expressions of numbers are odd numbers.

And therefore, any odd integer can be expressed in the form \[6q+1\], or \[6q+3\], or\[6q+5.\]

11. An Army Contingent of \[\mathbf{616}\] Members are to March Behind an Army Band of \[\mathbf{32}\] Members in a Parade. The Two Groups Are to March in the Same Number of Columns. What Is the Maximum Number of Columns in Which They Can March?

We have to find the \[HCF\left( 616,\text{ }32 \right)\] to find the maximum number of columns in which they can march. 

To find the HCF, we can use Euclid’s algorithm.

\[616=32\times 19+8\]

\[\Rightarrow 32=8\times 4+0\]

Hence,  \[HCF\left( 616,\text{ }32 \right)\] is \[8\].

Therefore, they can march in \[8\] columns each.

12. Use Euclid’s Division Lemma to Show That the Square of Any Positive Integer Is Either of Form \[\mathbf{3}m\]or \[\mathbf{3}m+\mathbf{1}\] for some integer \[m\]. 

[Hint: Let \[x\] be any positive integer then it is of the form\[\mathbf{3}q,\mathbf{3}q+\mathbf{1}\] or \[\mathbf{3}q+\mathbf{2}\text{ }\]. Now square each of these and show that they can be rewritten in the form \[\mathbf{3}m\]or \[\mathbf{3}m+\mathbf{1}\].]

Let\[\text{ }a\] be any positive integer and \[b=3\].

Then \[a=3q+r\] for some integer \[q\ge 0\]

And \[r=0,1,2\] because \[0\le r<3\]

Therefore, \[a=3q\] or \[3q+1\] or \[3q+2\]

\[\Rightarrow {{a}^{2}}={{(3q)}^{2}}\] or \[{{(3q+1)}^{2}}\] or \[{{(3q+2)}^{2}}\]

\[\Rightarrow {{a}^{2}}={{(9q)}^{2}}\] or \[9{{q}^{2}}+6q+1\] or \[9{{q}^{2}}+12q+4\]

\[\Rightarrow {{a}^{2}}=3\times {{(3q)}^{2}}\] or \[3(3{{q}^{2}}+2q)+1\] or \[3(3{{q}^{2}}+4q+1)+1\]

\[\Rightarrow a=3{{k}_{1}}\] or \[3{{k}_{2}}+1\] or \[3{{k}_{3}}+1\]

Where \[{{k}_{1}},{{k}_{2}},{{k}_{3}}\] are some positive integers 

Hence, it can be said that the square of any positive integer is either of the form \[3m\]or \[3m+1\].

Short Answer Questions (3 Marks)

1. Given \[\sec \theta =\dfrac{13}{12}\] , Calculate the Values for All Other Trigonometric Ratios. 

A triangle ABC right angled at B and angle A is 𝛳

Given:   \[\sec \theta =\dfrac{13}{12}\]

Let us consider a triangle \[ABC\] in which \[\angle A=\theta \] and \[\angle B={{90}^{\circ }}\]

Let \[AB=12k\] and \[AC=13k\]

Then, find the value of $BC$

\[BC=\sqrt{{{(AC)}^{2}}-{{(AB)}^{2}}}\]

\[\begin{align}&\Rightarrow\sqrt{{{(13k)}^{2}}-{{(12k)}^{2}}}\\&\Rightarrow \sqrt{69{{k}^{2}}-144{{k}^{2}}} \\ & \Rightarrow \sqrt{25{{k}^{2}}} \\ & \Rightarrow 5k \\ \end{align}\]

\[\therefore BC=5k\]

Since, \[\sec \theta =\dfrac{13}{12}\]

Similarly, 

\[\sin \theta =\dfrac{BC}{AC}=\dfrac{5k}{13k}=\dfrac{5}{13}\]

\[\cos \theta =\dfrac{AB}{AC}=\dfrac{12k}{13k}=\dfrac{12}{13}\]

\[\tan \theta =\dfrac{BC}{AB}=\dfrac{5k}{12k}=\dfrac{5}{12}\]

\[\cot \theta =\dfrac{AB}{BC}=\dfrac{12k}{5k}=\dfrac{12}{5}\]

\[\cos ec\theta =\dfrac{AC}{BC}=\dfrac{13k}{5k}=\dfrac{13}{5}\]

2. If \[\cot \theta =\dfrac{7}{8}\] , then Evaluate the Followings Equations :                                                                                        

i. \[\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

Given: \[\cot \theta =\dfrac{7}{8}\]

Then, \[AB=7k\] and \[BC=8k\]

Using Pythagoras theorem, find $AC$

\[AC=\sqrt{{{(BC)}^{2}}+{{(AB)}^{2}}}\]

\[\Rightarrow \sqrt{{{(8k)}^{2}}+{{(7k)}^{2}}}\]

\[\Rightarrow \sqrt{64{{k}^{2}}+49{{k}^{2}}}\]

\[\Rightarrow \sqrt{113{{k}^{2}}}\]

\[\Rightarrow \sqrt{113}k\]

\[\therefore AC=\sqrt{113}k\]

A triangle ABC right angled at B and angle A is 𝛳, sides AB=7k, BC=8k and AC=113k

Now find the value of trigonometric ratios.

\[\sin \theta =\dfrac{BC}{AC}=\dfrac{8k}{\sqrt{113}k}=\dfrac{8}{\sqrt{113}}\] and \[\cos \theta =\dfrac{AB}{AC}=\dfrac{7k}{\sqrt{113}k}=\dfrac{7}{\sqrt{113}}\].

\[\dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

We know that, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$

\[\Rightarrow \dfrac{1-{{\sin }^{2}}\theta }{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\Rightarrow \dfrac{49}{113}\cdot \dfrac{113}{64}\Rightarrow \dfrac{49}{64}\]

$\therefore \dfrac{(1+\sin \theta )(1-\sin \theta )}{(1+\cos \theta )(1-\cos \theta )}=\dfrac{49}{64}$

ii. \[{{\cot }^{2}}\theta \]

Given: \[{{\cot }^{2}}\theta \]

We know that, $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

\[\Rightarrow \dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{\dfrac{49}{113}}{\dfrac{64}{113}}\Rightarrow \dfrac{49}{64}\]

$\therefore {{\cot }^{2}}\theta =\dfrac{49}{64}$

Hence, $\dfrac{\left( 1+\sin \theta  \right)\left( 1-\sin \theta  \right)}{\left( 1+\cos \theta  \right)\left( 1-\cos \theta  \right)}$ and ${{\cot }^{2}}\theta $ are same.

3. If \[3\cot A=4\], then show that  \[\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\].              

A triangle ABC right angled at B and sides AB=4k, BC=3k and AC=5k

Given: \[3\cot A=4\]

Let us consider a triangle \[ABC\] in which  \[\angle B={{90}^{\circ }}\]

Then, \[3\cot A=4\] 

\[\Rightarrow \cot A=\dfrac{4}{3}\]

Let \[AB=4k\] and \[BC=3k\]

 \[\Rightarrow \sqrt{{{(3k)}^{2}}+{{(4k)}^{2}}}=\sqrt{16{{k}^{2}}+9{{k}^{2}}}\]

\[\Rightarrow \sqrt{25{{k}^{2}}}=5k\]

\[\therefore AC=5k\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{3k}{5k}=\dfrac{3}{5}\]  , \[\cos A=\dfrac{AB}{AC}=\dfrac{4k}{5k}=\dfrac{4}{5}\] and \[\tan A=\dfrac{BC}{AB}=\dfrac{3k}{4k}=\dfrac{3}{4}\].

To prove: \[\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\]

Let us take left-hand side 

L.H.S \[=\dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}\]

Substitute the value of $\tan A$.

\[\Rightarrow \dfrac{1-\dfrac{9}{16}}{1+\dfrac{9}{16}}=\dfrac{16}{25}-\dfrac{9}{25}\]

\[\Rightarrow \dfrac{7}{25}\]

\[\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}=\dfrac{7}{25}\]

R.H.S \[={{\cos }^{2}}A-{{\sin }^{2}}A\]

\[\Rightarrow {{\left( \dfrac{4}{5} \right)}^{2}}-{{\left( \dfrac{3}{5} \right)}^{2}}=\dfrac{16}{25}-\dfrac{9}{25}\]

$\therefore {{\cos }^{2}}A-{{\sin }^{2}}A=\dfrac{7}{25}$.

It shows that \[\text{L}\text{.H}\text{.S=R}\text{.H}\text{.S}\]

\[\therefore \dfrac{1-{{\tan }^{2}}A}{1+{{\tan }^{2}}A}={{\cos }^{2}}A-{{\sin }^{2}}A\]

4. In \[\Delta ABC\] Right Angles at \[B\], if \[A=\dfrac{1}{\sqrt{3}}\], then Find Value of the Following Equations:                                  

(i) \[\sin A\cos C+\cos A\sin C\]

A triangle ABC right angled at B and sides AB=3k, BC=k and AC=2k

Let \[BC=k\]and \[AB=\sqrt{3}k\]

Then, using Pythagoras theorem find $AC$

\[\Rightarrow \sqrt{{{(k)}^{2}}+{{(\sqrt{3}k)}^{2}}}\Rightarrow \sqrt{{{k}^{2}}+3{{k}^{2}}}\Rightarrow \sqrt{4{{k}^{2}}}\Rightarrow 2k\]

\[\therefore AC=2k\]

\[\sin A=\dfrac{BC}{AC}=\dfrac{k}{2k}=\dfrac{1}{2}\] and \[\cos A=\dfrac{AB}{AC}=\dfrac{\sqrt{3}k}{2k}=\dfrac{\sqrt{3}}{2}\]

For \[\angle C\], adjacent \[=BC\], opposite \[=AB\], and hypotenuse \[=AC\]

\[\sin C=\dfrac{AB}{AC}=\dfrac{\sqrt{3}k}{2k}=\dfrac{\sqrt{3}}{2}\] and \[\cos A=\dfrac{BC}{AC}=\dfrac{k}{2k}=\dfrac{1}{2}\]

Now find the values of the following equations,

\[\sin A\cos C+\cos A\sin C\]

\[\Rightarrow \dfrac{1}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{1}{4}+\dfrac{3}{4}\Rightarrow \dfrac{4}{4}\Rightarrow 1\]

\[\therefore \sin A\cos C+\cos A\sin C=1\]

(ii) \[\cos A\cos C-\sin A\sin C\]

\[\Rightarrow \dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}-\dfrac{1}{2}\times \dfrac{\sqrt{3}}{2}\Rightarrow \dfrac{\sqrt{3}}{4}-\dfrac{\sqrt{3}}{4}\Rightarrow 0\]

\[\therefore \cos A\cos C-\sin A\sin C=0\]

5. In \[\Delta PQR\], right angled at \[Q\],\[PR+QR=25cm\] and \[PQ=5cm\]. Determine the Values of  \[\sin P,\cos P\]  and \[\tan P\].                                                                      

A triangle PQR right angled at Q and sides PQ=5cm, QR= x cm and PR=(25-x) cm

Given: In \[\Delta PQR\], right angled at \[Q\]

And \[PR+QR=25cm\], \[PQ=5cm\]

Let us take \[QR=xcm\] and \[PR=(25-x)cm\]

By using Pythagoras theorem, find the value of $x$.

\[R{{P}^{2}}=R{{Q}^{2}}+Q{{P}^{2}}\]

\[\Rightarrow {{(25-x)}^{2}}={{(x)}^{2}}+{{(5)}^{2}}\Rightarrow 625-50x+{{x}^{2}}={{x}^{2}}+25\]

\[\Rightarrow -50x=-600\Rightarrow x=12\]

Hence, \[RQ=12cm\]and \[RP=25-12=13cm\]

Now, find the values of \[\sin P,\cos P\]   and \[\tan P\].     

\[\therefore \sin P=\dfrac{RQ}{RP}=\dfrac{12}{13}\], \[\cos P=\dfrac{PQ}{RP}=\dfrac{5}{13}\] and \[\tan P=\dfrac{RQ}{PQ}=\dfrac{12}{5}\].

6. If \[\tan (A+B)=\sqrt{3}\] and \[\tan (A-B)=\dfrac{1}{\sqrt{3}}\]; \[{{0}^{\circ }}<A+B\le {{90}^{\circ }}\]; \[A>B\]. Find \[A\]and \[B\].  

Given : $\tan \left( A+B \right)=\sqrt{3}$ and $\tan \left( A-B \right)=\dfrac{1}{\sqrt{3}}$.

We know that,$\tan 60{}^\circ =\sqrt{3}$ and $\tan 30{}^\circ =\dfrac{1}{\sqrt{3}}$.

$\tan \left( A+B \right)=\tan 60{}^\circ $

$\Rightarrow A+B=60{}^\circ $          …… (1)

$\tan \left( A-B \right)=\tan 30{}^\circ $

$\Rightarrow A-B=30{}^\circ $          …… (2)

Adding equation (1) and (2). We get,

$\begin{align} & A+B+A-B=60{}^\circ +30{}^\circ \\ & \Rightarrow 2A=90{}^\circ \Rightarrow A=45{}^\circ \\ \end{align}$

$\therefore A=45{}^\circ $

Put $A=45{}^\circ $ in equation (1).

$A+B=60{}^\circ $

$\Rightarrow 45{}^\circ +B=60{}^\circ \Rightarrow B=60{}^\circ -45{}^\circ \Rightarrow B=15{}^\circ $

$\therefore B=15{}^\circ $

Hence, $A=45{}^\circ $ and $B=15{}^\circ $.

7. Choose the Correct Option. Justify Your Choice:   

(i) \[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]\[=\]

\[1\] 

\[9\] 

\[8\] 

Ans: (B) $9$

\[9{{\sec }^{2}}A-9{{\tan }^{2}}A\]

\[\Rightarrow 9({{\sec }^{2}}A-{{\tan }^{2}}A)\Rightarrow 9\times 1\Rightarrow 9\]

(ii) \[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]\[=\]

\[0\] 

\[2\] 

none of these

Ans: (C) \[2\]

\[(1+\tan \theta +\sec \theta )(1+\cot \theta -\cos ec\theta )\]

\[\Rightarrow \left( 1+\dfrac{\sin \theta }{\cos \theta }+\dfrac{1}{\cos \theta } \right)\left( 1+\dfrac{\cos \theta }{\sin \theta }-\dfrac{1}{\sin \theta } \right)\Rightarrow \left( \dfrac{\cos \theta +\sin \theta +1}{\cos \theta } \right)\left( \dfrac{\sin \theta +\cos \theta -1}{\sin \theta } \right)\]

We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{(\cos \theta +\sin \theta )}^{2}}-{{(1)}^{2}}}{\cos \theta .\sin \theta }\Rightarrow \dfrac{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\]        

\[\Rightarrow \dfrac{1+2\cos \theta \sin \theta -1}{\cos \theta .\sin \theta }\Rightarrow \dfrac{2\cos \theta \sin \theta }{\cos \theta .\sin \theta }\Rightarrow 2\]

(iii) \[(\sec A+\tan A)(1-\sin A)\]\[=\]

\[\sec A\] 

\[\sin A\] 

\[\cos ecA\] 

Ans : (D) \[\cos A\]

\[(\sec A+\tan A)(1-\sin A)\]

\[\Rightarrow \left( \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A} \right)(1-\sin A)\Rightarrow \left( \dfrac{1+\sin A}{\cos A} \right)(1-\sin A)\]

We know that, \[1-{{\sin }^{2}}A={{\cos }^{2}}A\]

\[\Rightarrow \dfrac{1-{{\sin }^{2}}A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A}{\cos A}\Rightarrow \cos A\]  

(iv) \[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\]

\[{{\sec }^{2}}A\] 

\[-1\] 

\[{{\cot }^{2}}A\] 

Ans: (D) ${{\tan }^{2}}A$

\[\dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A}=\dfrac{{{\sec }^{2}}A-{{\tan }^{2}}A+{{\tan }^{2}}A}{\cos e{{c}^{2}}A-{{\cot }^{2}}A+{{\cot }^{2}}A}\]

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfac{rac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

\[\Rightarrow \dfr{{\sin }^{2}}A}{{{\cos }^{2}}A}\Rightarrow {{\tan }^{2}}A\].

Long Answer Questions (4 Marks)

1. Express the Trigonometric Ratios \[\sin A,\sec A\] and \[\tan A\] in Terms of \[\cot A\].  

Find the value for \[\sin A\] in terms of \[\cot A\]

By using identity \[\cos e{{c}^{2}}A-{{\cot }^{2}}A=1\]

$\therefore \sin A=\dfrac{1}{\sqrt{1+{{\cot }^{2}}A}}$

Express the value of \[\sec A\] in terms of \[\cot A\]

By using identity \[{{\sec }^{2}}A-{{\tan }^{2}}A=1\]

\[\Rightarrow {{\sec }^{2}}A=\dfrac{1+{{\cot }^{2}}A}{{{\cot }^{2}}A}\]

\[\Rightarrow \sec A=\dfrac{\sqrt{1+{{\cot }^{2}}A}}{\cot A}\]

Express the value of \[\tan A\] in terms of \[\cot A\]

We know that, $\tan A=\dfrac{1}{\cot A}$

\[\therefore \tan A=\dfrac{1}{\cot A}\]

2. Write the Other Trigonometric Ratios of \[A\] in Terms of \[\sec A\] .                 

Express the value of \[\sin A\] in terms of \[\sec A\]

By using identity, \[{{\sin }^{2}}A+{{\cos }^{2}}A=1\]

\[\Rightarrow {{\sin }^{2}}A=1-{{\cos }^{2}}A\Rightarrow 1-\dfrac{1}{{{\sec }^{2}}A}\Rightarrow \dfrac{{{\sec }^{2}}A-1}{{{\sec }^{2}}A}\]

Express the value of \[\cos A\] in terms of \[\sec A\]

We know that, \[\cos A=\dfrac{1}{\sec A}\]

\[\therefore \cos A=\dfrac{1}{\sec A}\]

Express the value of \[\tan A\] in terms of \[\sec A\]

$\therefore \tan A=\sqrt{{{\sec }^{2}}A-1}$

Express the value of \[\cos ecA\] in terms of \[\sec A\]

We know that, $\operatorname{cosec}A=\dfrac{1}{\sin A}$

Substitute the value of \[\sin A\]

\[\Rightarrow \cos ecA=\dfrac{1}{\dfrac{\sqrt{{{\sec }^{2}}A-1}}{\sec A}}\Rightarrow \dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}\]

$\therefore \operatorname{cosec}A=\dfrac{\sec A}{\sqrt{{{\sec }^{2}}A-1}}$

Express the value of \[\cot A\] in terms of \[\sec A\]

We know that, \[\cot A=\dfrac{1}{\tan A}\]

Substitute the value of \[\tan A\]

$\therefore \cot A=\dfrac{1}{\sqrt{{{\sec }^{2}}A-1}}$

3. Evaluate the following equations:                                                                                                       

(i). \[\dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]

We know that, \[\sin ({{90}^{\circ }}-\theta )=\cos \theta ,\,\cos ({{90}^{\circ }}-\theta )=\sin \theta \] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}({{90}^{\circ }}-{{63}^{\circ }})}{{{\cos }^{2}}({{90}^{\circ }}-{{73}^{\circ }})+{{\cos }^{2}}{{73}^{\circ }}}\]         

\[\Rightarrow \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\cos }^{2}}{{63}^{\circ }}}{{{\sin }^{2}}{{73}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}\]                   

\[\Rightarrow \dfrac{1}{1}\Rightarrow 1\]

$\therefore \dfrac{{{\sin }^{2}}{{63}^{\circ }}+{{\sin }^{2}}{{27}^{\circ }}}{{{\cos }^{2}}{{17}^{\circ }}+{{\cos }^{2}}{{73}^{\circ }}}=1$

(ii). \[\sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}\]

\[\Rightarrow \sin {{25}^{\circ }}\cos ({{90}^{\circ }}-{{25}^{\circ }})+\cos {{25}^{\circ }}\sin ({{90}^{\circ }}-{{25}^{\circ }})\]

\[\Rightarrow \sin {{25}^{\circ }}.\sin {{25}^{\circ }}+\cos {{25}^{\circ }}.\cos {{25}^{\circ }}\]                   

\[\Rightarrow {{\sin }^{2}}{{25}^{\circ }}+{{\cos }^{2}}{{25}^{\circ }}=1\]      

\[\therefore \sin {{25}^{\circ }}\cos {{65}^{\circ }}+\cos {{25}^{\circ }}\sin {{65}^{\circ }}=1\]

4. Choose the Correct Option. Justify Your Choice:  

\[\Rightarrow \dfrac{1-{{\sin }^{2}}A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A}{\cos A}\Rightarrow \cos A\] 

\[\Rightarrow \dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\Rightarrow \dfrac{\dfrac{1}{{{\cos }^{2}}A}}{\dfrac{1}{{{\sin }^{2}}A}}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{{{\cos }^{2}}A}\Rightarrow {{\tan }^{2}}A\].

5. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                      

(i). ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

Given: ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

We know that, \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], $\operatorname{cosec}\theta =\dfrac{1}{\sin \theta }$ and $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$

Then, let us take left-hand side

\[\Rightarrow {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}={{\operatorname{cosec}}^{2}}\theta +{{\cot }^{2}}\theta -2\operatorname{cosec}\theta \cot \theta \] 

\[\Rightarrow \dfrac{1}{{{\sin }^{2}}\theta }+\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-2\times \dfrac{1}{\sin \theta }.\dfrac{\cos \theta }{\sin \theta }\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }-\dfrac{2\cos \theta }{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{1+{{\cos }^{2}}\theta -2\cos \theta }{{{\sin }^{2}}\theta }\Rightarrow \dfrac{{{(1-\cos \theta )}^{2}}}{{{\sin }^{2}}\theta }\]

\[\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{1-{{\cos }^{2}}\theta }\Rightarrow \dfrac{(1-\cos \theta )(1-\cos \theta )}{(1+\cos \theta )(1-\cos \theta )}\]

\[\Rightarrow \dfrac{1-\cos \theta }{1+\cos \theta }=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta }{1+\cos \theta }$

(ii) \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

Given: \[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A\]

\[\dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{{{\cos }^{2}}A+1+{{\sin }^{2}}A+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{{{\cos }^{2}}A+{{\sin }^{2}}A+1+2\sin A}{(1+\sin A)\cos A}\]      

\[\Rightarrow \dfrac{1+1+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2+2\sin A}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2(1+\sin A)}{(1+\sin A)\cos A}\]

\[\Rightarrow \dfrac{2}{\cos A}\Rightarrow 2\sec A=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A}{1+\sin A}+\dfrac{1+\sin A}{\cos A}=2\sec A$

(iii). \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Given: \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

We know that, \[{{a}^{3}}-{{b}^{3}}=(a-b)({{a}^{2}}+{{b}^{2}}+ab)\] and \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]

Then, let us take L.H.S

\[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=\dfrac{\dfrac{\sin \theta }{\cos \theta }}{1-\dfrac{\cos \theta }{\sin \theta }}+\dfrac{\dfrac{\cos \theta }{\sin \theta }}{1-\dfrac{\sin \theta }{\cos \theta }}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }\times \dfrac{\sin \theta }{\sin \theta -\cos \theta }+\dfrac{\cos \theta }{\sin \theta }\times \dfrac{\cos \theta }{\cos \theta -\sin \theta }\]

\[\Rightarrow \dfrac{{{\sin }^{2}}\theta }{\cos \theta (\sin \theta -\cos \theta )}-\dfrac{{{\cos }^{2}}\theta }{\sin \theta (\sin \theta -\cos \theta )}\]

\[\Rightarrow \dfrac{{{\sin }^{3}}\theta -{{\cos }^{3}}\theta }{\sin \theta \cos \theta (\sin \theta -\cos \theta )}\]                     

\[\Rightarrow \dfrac{1+\sin \theta \cos \theta }{\sin \theta \cos \theta }\]

\[\Rightarrow \dfrac{1}{\sin \theta \cos \theta }+1\]

\[\Rightarrow 1+\dfrac{1}{\sin \theta \cos \theta }\Rightarrow 1+\sec \theta \cos ec\theta \] = R.H.S

$\therefore \dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta $

(iv) \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Given: \[\dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}\]

Then, let us take L.H.S 

\[\dfrac{1+\sec A}{\sec A}=\dfrac{1+\dfrac{1}{\cos A}}{\dfrac{1}{\cos A}}\]

\[\Rightarrow \dfrac{\cos A+1}{\cos A}\times \dfrac{\cos A}{1}\Rightarrow 1+\cos A\]

\[\Rightarrow 1+\cos A\times \dfrac{1-\cos A}{1-\cos A}\Rightarrow \dfrac{1-{{\cos }^{2}}A}{1-\cos A}\]

\[\Rightarrow \dfrac{{{\sin }^{2}}A}{1-\cos A}=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{1+\sec A}{\sec A}=\dfrac{{{\sin }^{2}}A}{1-\cos A}$

(v) \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\], using the identity \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

Given: \[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A\]

We know that, \[\cos e{{c}^{2}}A=1+{{\cot }^{2}}A\]

\[\dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}\]

Dividing all terms by \[\sin A\]

\[\Rightarrow \dfrac{\cot A-1+\cos ecA}{\cot A+1-\cos ecA}\]

\[\Rightarrow \dfrac{\cot A+\cos ecA-1}{\cot A-\cos ecA+1}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)-(\cos e{{c}^{2}}A-{{\cot }^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)+({{\cot }^{2}}A-\cos e{{c}^{2}}A)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \dfrac{(\cot A+\cos ecA)(1+\cot A-\cos ecA)}{(1+\cot A-\cos ecA)}\]

\[\Rightarrow \cot A+\cos ecA=\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\cos A-\sin A+1}{\cos A+\sin A-1}=\cos ecA+\cot A$

(vi) \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

Given: \[\sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A\]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \] and \[(a+b)(a-b)={{a}^{2}}-{{b}^{2}}\]

\[\sqrt{\dfrac{1+\sin A}{1-\sin A}}\]

Let us take conjugate of the term. Then,

\[\Rightarrow \sqrt{\dfrac{1+\sin A}{1-\sin A}}\times \sqrt{\dfrac{1+\sin A}{1+\sin A}}\]

\[\Rightarrow \sqrt{\dfrac{{{(1+\sin A)}^{2}}}{1-{{\sin }^{2}}A}}\]

\[\Rightarrow \sqrt{\dfrac{{{\left( 1+\sin A \right)}^{2}}}{{{\operatorname{Cos}}^{2}}A}}\]

\[\Rightarrow \dfrac{1+\sin A}{\cos A}\Rightarrow \dfrac{1}{\cos A}+\dfrac{\sin A}{\cos A}\]                           

\[\Rightarrow \sec A+\tan A=\text{R}\text{.H}\text{.S}\]           

$\therefore \sqrt{\dfrac{1+\sin A}{1-\sin A}}=\sec A+\tan A$       

Hence proved.        

(vii) \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

Given: \[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta \]

We know that, \[1-{{\sin }^{2}}\theta ={{\cos }^{2}}\theta \]

\[\dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2{{\cos }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta \left[ 2(1-{{\sin }^{2}}\theta )-1 \right]}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\sin }^{2}}\theta )}{\cos \theta (2-2{{\sin }^{2}}\theta -1)}\]

\[\Rightarrow \dfrac{\sin \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}{\cos \theta (1-2{{\operatorname{Sin}}^{2}}\theta )}\]

\[\Rightarrow \dfrac{\sin \theta }{\cos \theta }=\tan \theta =\text{R}\text{.H}\text{.S}\]

$\therefore \dfrac{\sin \theta -2{{\sin }^{3}}\theta }{2{{\cos }^{3}}\theta -\cos \theta }=\tan \theta $

(viii) \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

Given: \[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A\]

We know that, \[\cos e{{c}^{2}}\theta =1+{{\cot }^{2}}\theta \] and \[{{\sec }^{2}}\theta =1+{{\tan }^{2}}\theta \]

\[{{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}={{\left( \sin A+\dfrac{1}{\sin A} \right)}^{2}}+{{\left( \cos A+\dfrac{1}{\cos A} \right)}^{2}}\]

\[={{\sin }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+2\sin A.\dfrac{1}{\sin A}+{{\cos }^{2}}A+\dfrac{1}{{{\cos }^{2}}A}+2\cos A.\dfrac{1}{\cos A}\]

\[=2+2+{{\sin }^{2}}A+{{\cos }^{2}}A+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=4+1+\dfrac{1}{{{\sin }^{2}}A}+\dfrac{1}{{{\cos }^{2}}A}\]

\[=5+\cos e{{c}^{2}}A+{{\sec }^{2}}A\]

\[=5+1+{{\cot }^{2}}A+1+{{\tan }^{2}}A\]

\[=7+{{\tan }^{2}}A+{{\cot }^{2}}A=\text{R}\text{.H}\text{.S}\]

$\therefore {{(\sin A+\cos ecA)}^{2}}+{{(\cos A+\sec A)}^{2}}=7+{{\tan }^{2}}A+{{\cot }^{2}}A$

(ix) \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

Given: \[(\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}\]

\[(\cos ecA-\sin A)(\sec A-\cos A)=\left( \dfrac{1}{\sin A}-\sin A \right)\left( \dfrac{1}{\cos A}-\cos A \right)\]

\[=\left( \dfrac{1-{{\sin }^{2}}A}{\sin A} \right)\left( \dfrac{1-{{\cos }^{2}}A}{\cos A} \right)\]

\[=\dfrac{{{\cos }^{2}}A}{\sin A}\times \dfrac{{{\sin }^{2}}A}{\cos A}\]

\[=\sin A.\cos A\]

\[=\dfrac{\sin A.\cos A}{{{\sin }^{2}}A+{{\cos }^{2}}A}\]

Dividing all the terms by \[\sin A.\cos A\]

\[=\dfrac{\dfrac{\sin A.\cos A}{\sin A.\cos A}}{\dfrac{{{\sin }^{2}}A}{\sin A.\cos A}+\dfrac{{{\cos }^{2}}A}{\sin A.\cos A}}\]

\[=\dfrac{1}{\dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}}\]

\[=\dfrac{1}{\tan A+\cot A}=\text{R}\text{.H}\text{.S}\]

$\therefore (\cos ecA-\sin A)(\sec A-\cos A)=\dfrac{1}{\tan A+\cot A}$

(x) \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

Given:\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

We know that, \[1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta \] and \[1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}A\]

\[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)=\dfrac{{{\sec }^{2}}A}{\cos e{{c}^{2}}A}\]

\[=\dfrac{1}{{{\cos }^{2}}A}\times \dfrac{{{\sin }^{2}}A}{1}\]

\[={{\tan }^{2}}A=\text{R}\text{.H}\text{.S}\]

Now, prove the Middle side 

\[{{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\left( \dfrac{1-\tan A}{1-\dfrac{1}{\tan A}} \right)}^{2}}\]

\[={{\left( \dfrac{1-\tan A}{\dfrac{\tan A-1}{\tan A}} \right)}^{2}}\]

\[=\left( \dfrac{1-\tan A}{\dfrac{-(1-\tan A)}{\tan A}} \right)\]

\[={{(-\tan A)}^{2}}\]

$\therefore \left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A$

6. Use Euclid’s Division Algorithm to Find the \[\text{HCF}\] of:                                     

(i) \[\mathbf{135}\] and \[\mathbf{225}\]

Given: \[135\] and \[225\]

We have \[225>135\],

So, we have to apply the division lemma to \[225\] and \[135\] to obtain

\[225=135\times 1+90\]

Here remainder\[90\ne 0\], again we are applying the division lemma to \[135\] and \[90\] to obtain 

\[135=90\times 1+45\]

Again, the remainder\[45\ne 0\], then apply the division lemma to obtain 

\[90=2\times 45+0\]

Since now we got remainder as zero. Here, the process get stops. 

The divisor at this stage is \[45\]

Therefore, the HCF of \[135\] and \[225\] is \[45\].

(ii) \[\mathbf{196}\] and\[~\mathbf{38220}\] 

Given: \[196\] and \[38220\]

We have \[38220>196\], 

So, we have to apply the division lemma to \[38220\] and \[196\] to obtain 

\[38220=196\times 195+0\]

Since we get the remainder as zero, the process stops here. 

The divisor at this stage is \[196\], 

Therefore, HCF of \[196\] and \[38220\] is \[196\].  

(iii) \[~\mathbf{867}\] and \[\mathbf{255}\]

Given: \[867\] and \[255\]

We have 867 > 255, 

So, we have to apply the division lemma to \[867\]and \[255\] to obtain 

\[867=255\times 3+102\]

Here remainder \[102\ne 0\], again apply the division lemma to\[255\] and \[102\] to obtain 

\[255=102\times 2\text{ }51\]

Again, remainder \[51\ne 0\], again apply the division lemma to \[102\] and \[51\] to obtain 

\[102=51\times 2+0\]

The divisor at this stage is \[51\], 

Therefore, HCF of \[867\] and \[255\] is \[51\].

7. Evaluate the Following Equations:                                                                                                              

i. \[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

Given: \[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

We know that, $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2},\,\sin 30{}^\circ =\dfrac{1}{2}$ and $\cos 60{}^\circ =\dfrac{1}{2},\,\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

\[\sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}\]

\[=\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}\times \dfrac{1}{2}\]

\[=\dfrac{3}{4}+\dfrac{1}{4}\]

\[=\dfrac{4}{4}=1\]

\[\therefore \sin {{60}^{\circ }}\cos {{30}^{\circ }}+\sin {{30}^{\circ }}\cos {{60}^{\circ }}=1\]

ii. \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

Given: \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

We know that, $\tan 45{}^\circ =1$, $\sin 60{}^\circ =\dfrac{\sqrt{3}}{2}$ and $\cos 30{}^\circ =\dfrac{\sqrt{3}}{2}$

Then, \[2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}\]

\[=2{{(1)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}\]

\[=2+\dfrac{3}{4}-\dfrac{3}{4}=2\]

$\therefore 2{{\tan }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\sin }^{2}}{{60}^{\circ }}=2$

iii. \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}\]

Given: \[\dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}\]

We know that, $\cos 45{}^\circ =\dfrac{1}{\sqrt{2}},\,\sec 30{}^\circ =\dfrac{2}{\sqrt{3}}$ and $\operatorname{cosec}30{}^\circ =2$

\[=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2}{\sqrt{3}}+2}\]

\[=\dfrac{\dfrac{1}{\sqrt{2}}}{\dfrac{2+2\sqrt{3}}{\sqrt{3}}}\]

\[=\dfrac{1}{\sqrt{2}}\times \dfrac{\sqrt{3}}{2+2\sqrt{3}}\]

\[=\dfrac{\sqrt{3}}{\sqrt{2}\times 2(\sqrt{3}+1)}\]

\[=\dfrac{\sqrt{3}}{\sqrt{2}\times 2(\sqrt{3}+1)}\times \dfrac{\sqrt{3}-1}{\sqrt{3}-1}\]

\[=\dfrac{\sqrt{3}\left( \sqrt{3}-1 \right)}{\sqrt{2}\times 2(3-1)}\]

\[=\dfrac{\sqrt{3}(\sqrt{3}-1)}{4\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}\]\[\]

\[=\dfrac{3\sqrt{2}-\sqrt{6}}{8}\]

$\therefore \dfrac{\cos {{45}^{\circ }}}{\sec {{30}^{\circ }}+\cos ec{{30}^{\circ }}}=\dfrac{3\sqrt{2}-\sqrt{6}}{8}$

(iv) \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

Given: \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

We know that,  $\sin 30{}^\circ =\dfrac{1}{2},\,\tan 45{}^\circ =1,\,\operatorname{cosec}60{}^\circ =\dfrac{2}{\sqrt{3}},\,\sec 30{}^\circ =\dfrac{2}{\sqrt{3}},\,\cos 60{}^\circ =\dfrac{1}{2}$ and $\cot 45{}^\circ =1$.

Then, \[\dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}\]

$=\dfrac{{1}/{2}\;+1-{2}/{\sqrt{3}}\;}{{2}/{\sqrt{3}}\;+{1}/{2}\;+1}$

\[=\dfrac{\dfrac{\sqrt{3}+2\sqrt{3}-4}{2\sqrt{3}}}{\dfrac{4+\sqrt{3}+2\sqrt{3}}{2\sqrt{3}}}\]

\[=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\]

\[=\dfrac{3\sqrt{3}-4}{3\sqrt{3}+4}\times \dfrac{3\sqrt{3}-4}{3\sqrt{3}-4}\]

\[=\dfrac{27+16-24\sqrt{3}}{27-16}\]

\[=\dfrac{43-24\sqrt{3}}{11}\]

$\therefore \dfrac{\sin {{30}^{\circ }}+\tan {{45}^{\circ }}-\cos ec{{60}^{\circ }}}{\sec {{30}^{\circ }}+\cos {{60}^{\circ }}+\cot {{45}^{\circ }}}=\dfrac{43-24\sqrt{3}}{11}$

(v) \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}\]

Given: \[\dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}\]. Then,

\[=\dfrac{5{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{3}{\sqrt{3}} \right)}^{2}}-{{(1)}^{2}}}{{{\left( \dfrac{1}{2} \right)}^{2}}+{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}}\]

\[=\dfrac{5\times \dfrac{1}{4}+4\times \dfrac{4}{3}-1}{\dfrac{1}{4}+\dfrac{3}{4}}\]

\[=\dfrac{\dfrac{1}{12}\times 67}{\dfrac{4}{4}}\]

\[=\dfrac{67}{12}\]

$\therefore \dfrac{5{{\cos }^{2}}{{60}^{\circ }}+4{{\sin }^{2}}{{30}^{\circ }}-{{\tan }^{2}}{{45}^{\circ }}}{{{\sin }^{2}}{{30}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}}=\dfrac{67}{12}$

8. Prove the Following Identities, Where the Angles Involved are Acute Angles for Which the Expressions are Defined:                                      

(i) ${{(\operatorname{cosec}\theta -\cot \theta )}^{2}}=\dfrac{1-\cos \theta  }{1+\cos \theta }$

(iii) \[\dfrac{\tan \theta }{1-\cot \theta }+\dfrac{\cot \theta }{1-\tan \theta }=1+\sec \theta \operatorname{cosec}\theta \]

Given: \[\left( \dfrac{1+{{\tan }^{2}}A}{1+{{\cot }^{2}}A} \right)={{\left( \dfrac{1-\tan A}{1-\cot A} \right)}^{2}}={{\tan }^{2}}A\]

Key Topics Covered in Chapter 8

Chapter 8 of Class 10 Mathematics consists of the discussion of the following key concepts.

Basic Trigonometry

Opposite & Adjacent Sides In A Right-Angled Triangle

Basic Trigonometric Ratios

Standard Values Of Trigonometric Ratios 

Complementary Trigonometric Ratios 

Significance of Important Questions of Trigonometry Class 10

Trigonometry is one of the most significant parts of Important Questions for Class 10 Maths Chapter 8. This chapter revolves around the memorization and conceptual understanding of the user’s problem-solving ability of a given space, preferably a triangle. These Class 10 Maths Trigonometry Important Questions help students to have a better understanding and application of trigonometry in the real world and clear the base around its introductory stage, to help students retain its benefits in the long run.

Did You Know?

Trigonometry has different applications and a few of them are mentioned below:

It is used in cartography for the creation of maps.

It has applications in the aviation industry and satellite systems.

It is also used to describe light and sound waves. 

This was the complete discussion on CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry important questions. If you are a Class 10 student we highly recommend downloading and practising the important questions on trigonometry. Practising these important questions will not only help you understand the concepts but will also help you in analysing the important topics and exam patterns. 

We wish you all the very best! Be exam ready with Vedantu!

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FAQs on Important Questions for CBSE Class 10 Maths Chapter 8 Introduction to Trigonometry

1. Is Studying Chapter 8 Maths Class 10 Important Questions Helpful in Preparing for the Exams?

Studying Trigonometry Important Questions Class 10 helps you uncover the different aspects of trigonometry and teach the subject at its core. The pool of experienced teachers who have a complete idea of the marking schematics and all the necessary questions that can come in the exams are focused on providing the students with a complete kit that helps them secure maximum marks in their upcoming exams. Further, the step by step explanation of practical problems is what set our solutions apart, making it completely reliable a solution to start preparing for your exams.

2. What are the important questions in Chapter 8 Trigonometry of Class 10 Maths?

Class 10 Maths Chapter 8 is related to trigonometry and its functions. Important questions given on Vedantu for trigonometry Class 10 Chapter 8 can help students to prepare for their exams. Important questions are given according to the latest syllabus and pattern of CBSE. All questions are prepared using sample papers, previous year papers, and the latest trends in the CBSE board exams. Students can use important questions in trigonometry Class 10 to prepare for their final board exams. These solutions are available on Vedantu's official website(vedantu.com) and mobile app free of cost.

3. What is the easiest way to solve Chapter 8 Trigonometry of Class 10 Maths questions?

Students have to understand the basic concepts of Trigonometry for solving Class 10 Maths Chapter 8 questions. They have to memorize the different identities properly for solving the NCERT questions quickly. Students can also refer to the important questions in Class 10 Maths Chapter 8 available for free on Vedantu to understand the basic concepts and formulas used for solving trigonometry easily.  They can visit Vedantu to download important questions for Class 10 Maths Chapter 8 for regular practice.

4. How many exercises are there in Chapter 8 Trigonometry of Class 10 Maths?

In Trigonometry Class 10 Maths Chapter 8, there are four exercises. All exercises are based on different concepts related to trigonometry. Students need to solve all questions given in four exercises in the NCERT book to understand the concepts of trigonometry and for scoring high marks in Class 10 maths board exams. Students can practice solving trigonometry from the NCERT Solutions available on Vedantu in the offline and online mode. They can download the NCERT solutions for Class 10 Maths Chapter 8 on computers to work offline.

5. What are the different topics studied in Chapter 8 Trigonometry of Class 10 Maths?

The different topics studied in Class 10 Maths Chapter 10 include an introduction to trigonometry,  information about trigonometric ratios, trigonometric ratios of specific angles, trigonometric ratios of complementary angles, and trigonometric identities. Students will learn all topics step by step in each exercise of the chapter. They can practice NCERT Solutions given on different topics of trigonometry on Vedantu. NCERT Solutions for trigonometry Class 10 are prepared by experts to help students understand the concepts and score high marks. 

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Chapter 8 Class 10 Introduction to Trignometry

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The chapter is updated according to the new NCERT, for 2023-2024 Board Exams.

Get NCERT Solutions with videos of all questions and examples of Chapter 8 Class 10 Trigonometry. Videos of all questions are made with step-by-step explanations. Check it out now.

Trigonometry means studying relationship between measures of triangle. Usually, we talk about right triangles when we study trigonometry,

In this chapter, we will study

  • What is sin, cos, tan ( Sine, cosine, tangent) ... and how they are found in a triangle
  • What is sec, cosec, cot, and how is it related to sin, cos, tan.
  • (Sin, cos, tan, sec, cosec, cot are known as Trigonometric Ratios)
  • Then, we study Trigonometric ratios of specific angles l ike 0°, 30°, 45°, 60°, 90° ; and do some questions
  • We study the formulas of sin (90 - θ) , cos (90 - θ), tan (90 - θ)
  • And then we study Trigonometric Identities, and how other identities are derived from sin 2 θ + cos 2 θ = 1

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Important Questions for Class 10 Maths Chapter 8 Introduction to Trigonometry

Chapter 8 introduction to trigonometry important questions for cbse class 10 maths board exams.

Chapter 8 Introduction to Trigonometry Important Questions for CBSE Class 10 Maths Board Exams

Important Questions for Chapter 8 Introduction to Trigonometry Class 10 Maths

Introduction to trigonometry class 10 maths important questions very short answer (1 mark).

We know that, sec 2 θ – tan 2 θ = 1 ⇒ (sec θ + tan θ) (sec θ – tan θ) = 1 ⇒  (7) (sec θ – tan θ) = 1 …[sec θ + tan θ = 7] ∴ sec θ – tan θ = 1/7

2. Prove that

(1+tan A - sec A) (1+tan A + sec A) = 2 tan A

LHS = (1+tan A) 2 - sec 2 A

= 1+ tan 2 A + 2 tan A - sec 2 A

= sec 2 A + 2 tan A - sec2 A

= 2 tan A = RHS

3. If tan A = cot B, then find the value of (A+B)

tan A = cot B

tan A = tan(90°-B)

A = 90° - B

Thus, A + B = 90°

4. If sinθ + sin 2 θ = 1 then prove that cos 2 θ + cos 4 θ = 1.

sinθ + sin 2 θ = 1

⇒ sinθ + (1-cos 2 θ) = 1

⇒ sinθ - cos 2 θ = 0

⇒ sinθ = cos 2 θ

Squaring both sides, we get

sin 2 θ = cos 4 θ

⇒ 1 - cos 2 θ = cos 4 θ

⇒ cos 4 θ + cos 2 θ = 1

5. If tan θ + cot θ = 5, find the value of tan2θ + cotθ.

tan θ + cot θ = 5 …[Given] ⇒ tan 2 θ + cot 2 θ + 2 tan θ cot θ = 25 …[Squaring both sides] ⇒ tan 2 θ + cot 2 θ + 2 = 25 ∴ tan 2 θ + cot 2 θ = 23

6. If sec 2A = cosec (A – 27°) where 2A is an acute angle, find the measure of ∠A.

sec 2A = cosec (A – 27°) ⇒ cosec(90° – 2A) = cosec(A – 27°) …[∵ sec θ = cosec (90° – θ)] ⇒ 90° – 2A = A – 27° ⇒ 90° + 27° = 2A + A ⇒ 3A = 117° ∴ ∠A = 117°/3 = 39°

7. Evaluate: sin 2 19° + sin 2 71°.

sin 2 19° + sin 2 71° = sin 2 19° + sin 2 (90° – 19°) …[∵ sin(90° – θ) = cos θ] = sin 2 19° + cos 2 19° = 1 …[∵ sin 2 θ + cos 2 θ = 1]

8. In a triangle ABC, write cos(B+C/2) in terms of angle A.

trigonometry tough questions class 10

sin 2 41° + sin 2 49°

= sin 2 (90°-49°) + sin 2 49°

= cos 2 49° + sin 2 49°

11. If tan A = cot B, prove that A + B = 90°.

tan A = cot B ∴ tan A = tan (90° − B) ⇒ A = 90° − B ⇒ A + B = 90°

12. Express sin 67° + cos 75° in terms of ratios of angles between 0° and 45°.

 Solution

∵ 67° = 90° − 23° and 75° = 90° − 15° 

∴ sin 67° + cos 75°

= sin (90° − 23°) + cos (90° − 15°)

= cos 23° + sin 15°

13. What is the value of sinθ. cos(90° - θ) + cosθ . sin(90° - θ)?

sinθ ·cos(90° − θ) + cosθ · sin(90° − θ)

= sinθ · sinθ + cosθ · cosθ  [∵ cos(90° − θ) = sinθ , sin(90° − θ) = cos θ] = sin 2 θ + cos 2 θ = 1

14. If tan θ = cot (30° + θ ), find the value of θ .

We have, tan θ = cot (30° + θ) = tan [90° − (30° + θ)] = tan [90° − 30° − θ] = tan (60° − θ) ⇒ θ = 60° − θ ⇒ θ + θ = 60° ⇒ 2θ = 60° ⇒ θ = 60°/2 ⇒ θ = 30°

15. If sin 3θ = cos (θ - 6)° and 3θ and (θ - 6)° are acute angles, find the value of θ.

We have, sin3θ = cos(θ − 6)° = sin[90°−(θ − 6)°] ∵ [sin (90° − θ) = cos θ] ⇒ 3θ = 90° − (θ − 6)° ⇒ 3θ = 90° − θ + 6° ⇒ 3θ + θ = 96° ⇒ 4θ = 96°/4 ⇒ θ = 24°

16. Show that: tan 10° tan 15° tan 75° tan 80° = 1

L.H.S. = tan 10° tan 15° tan 75° tan 80° = tan (90° − 80°) tan 15° tan (90° − 15°) tan 80° = cot 80° tan 15 cot 15° tan 80° = (cot 80° × tan 80°) × (tan 15° × cot 15°) = 1× 1 = 1 = R.H.S.

Introduction to Trigonometry Class 10 Maths Important Questions Short Answer-I (2 Marks)

trigonometry tough questions class 10

tan 4 θ + tan 2 θ = tan 2 θ (1+ tan 2 θ)

= tan 2 θ × sec 2 θ

= (sec 2 θ - 1) sec 2 θ

= sec 4 θ - sec 2 θ

21. Find the value of (sin 2 33° + sin 2 57°)

sin 2 33° + sin 2 57° ⇒ sin 2 33° + sin 2 ( 90° - 33°) ⇒ sin 2 33° + cos 2 33 [Using sin(90° - θ) = cos θ] ⇒ 1 [Using sin 2 θ + cos 2 θ = 1]

22. If ∆ABC is right angled at B, what is the value of sin (A + C).

∠B = 90° …[Given] ∠A + ∠B + ∠C = 180° …[Angle sum property of a ∆] ⇒ ∠A + ∠C + 90° = 180° ⇒ ∠A + ∠C = 90° ∴ sin (A + C) = sin 90° = 1 …(taking sin both side)

23. If tan θ = a/x, find the value of x/√a 2 +x 2

trigonometry tough questions class 10

L.H.S. = x 2 – y 2 = (p sec θ + q tan θ) 2 – (p tan θ + q sec θ) 2 = p 2 sec 2 θ + q 2 tan 2 θ + 2 pq secθ tanθ - (p 2 tan 2 θ + q 2 sec 2 θ + 2pq secθ tanθ) = p 2 sec 2 θ + 2 tan 2 θ + 2pq secθ tanθ – p 2 tan 2 θ – q 2 sec 2 θ – 2pq secθ tanθ = p 2 (sec 2 θ – tan 2 θ) – q 2 (sec 2 θ – tan 2 θ) = p 2 – q 2 …[sec 2 θ – tan 2 θ = 1] = R.H.S.

25. If x = a cos θ – b sin θ and y = a sin θ + b cos θ, then prove that a 2 + b 2 = x 2 + y 2 .

R.H.S. = x 2 + y 2 = (a cos θ – b sin θ) 2 + (a sin θ + b cos θ) 2 = a 2 cos 2 θ + b 2 sin 2 θ – 2ab cosθ sinθ + a 2 sin 2 θ + b 2 cos 2 θ + 2ab sinθ cosθ = a 2 (cos 2 θ + sin 2 θ) + b 2 (sin 2 θ + cos 2 θ) = a 2 + b 2 = L.H.S. …[∵ cos 2 θ + sin 2 θ = 1]

cot 75° + cosec 75° = cot(90° – 15°) + cosec(90° – 15°) = tan 15° + sec 15° …[cot(90°-A) = tan A] = cosec(90° – A) = sec A

27. If cos (A + B) = 0 and sin (A – B) = 3, then find the value of A and B where A and B are acute angles.

trigonometry tough questions class 10

Putting the value of B in (i), we get

⇒ A = 30° + 30° = 60°

∴ A = 60°, B = 30°

28. If sin (A + 2B) = √3/2 and cos (A + 4B) = 0, A > B, and A + 4B ≤ 90° then find A and B.

sin (A + 2B) = √3/2 ⇒ sin(A + 2B) = sin 60° Hence, A + 2B = 60° ...(i) Also, we have cos(A + 4B) = 0 ⇒ cos(A + 4B) = cos 90° ⇒ A + 4B = 90° ...(ii) Subtracting (ii) from (i), we have -2B = - 30° ⇒ B = 15° Put B = 15° in eq. (i), we have A + 2(15°) = 60° ⇒ A + 30° = 60° ⇒ A = 30°

Introduction to Trigonometry Class 10 Maths Important Questions Short Answer-II (3 Marks)

29. Prove that:

tanθ/1-tanθ - cotθ/1-cotθ = cosθ+sinθ/cosθ-sinθ

trigonometry tough questions class 10

cos x = cos 40° sin 50° + sin 40° cos 50° ⇒ cos x = cos 40° sin(90° – 40°) + sin 40°.cos(90° – 40°) ⇒ cos x = cos 2 40° + sin 2 40° ⇒ cos x = 1 …[∵ cos 2 A + sin 2 A = 1] ⇒ cos x = cos 0° ⇒ x = 0°

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Class 10 Maths Chapter 18 Trigonometry Important Questions

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In Class 10 ICSE (Indian Certificate of Secondary Education) mathematics, Chapter 2 may not specifically cover the topic of "Banking" as its primary focus. Instead, the curriculum typically includes topics related to commercial mathematics, which can encompass banking concepts. Commercial mathematics includes areas such as simple interest, compound interest, and profit and loss, which are essential aspects of banking and finance. Here are trigonometric identities class 10 ICSE important questions.

Introduction

What is trigonometry, class 10 trigonometry important questions and answers, icse class 10 maths chapter wise important questions, frequently asked questions.

In Class 10 ICSE (Indian Certificate of Secondary Education) mathematics, the chapter on "Trigonometry" is a fundamental topic that explores the relationships between angles and sides of right triangles. Trigonometry is a branch of mathematics that has wide-ranging applications, including in navigation, engineering, physics, and more. Here's an introduction to trigonometry in Class 10 ICSE mathematics: Importance of Trigonometry: Trigonometry is a critical branch of mathematics because it helps us solve real-world problems involving angles and distances. It's especially valuable in fields that require precise measurements and calculations. Common Trigonometric Formulas: Trigonometric Identities: These include the Pythagorean identities (sin^2θ + cos^2θ = 1) and various angle sum and difference identities. Special Angles: Trigonometry deals with special angles like 30 degrees, 45 degrees, and 60 degrees, which have well-defined trigonometric values.

Ans: Trigonometry is the study of the relationships between the angles and sides of triangles. It primarily focuses on right triangles, which have one angle equal to 90 degrees. Trigonometric functions and ratios, such as sine, cosine, and tangent, are used to establish these relationships. Key Concepts and Objectives: Trigonometric Ratios: The fundamental trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). They relate the angles of a right triangle to the lengths of its sides. Right Triangles: Trigonometry primarily deals with right triangles, where one angle is 90 degrees. The side opposite the right angle is the hypotenuse, and the other two sides are the legs. Sine, Cosine, and Tangent: These trigonometric ratios are defined as follows: Sine (sin): Opposite side / Hypotenuse Cosine (cos): Adjacent side / Hypotenuse Tangent (tan): Opposite side / Adjacent side

trigonometry class 10 icse important questions

Q1. \(\frac{cos A}{1-sin A}\) - tan A =

(a) cos a (b) sec a (c) sin a (d) cosec a.

Explanation: \(\frac{cos A}{1- sin A}\) - tan A = \(\frac{cos A(1+sin A)}{1-sin A(1+ sin A)}\)- tan A =\(\frac{cos A(1+ sin A)}{(1+sin^2 A)}\)-tan A =\(\frac{cos A(1+ sin A)}{cos^2 A}\)- tan A =\(\frac{(1+sin A)}{cos A}\)-tan A =\(\frac{1}{cos A}\)+\(\frac{sin A}{cos A}\)- tan A

Q2. \(\frac{cos A}{1-sin A}\)-tan A=

(a) cos a (b) sec a (c) sin a (d) cosec a.

Ans . (b) Explanation: \(\frac{cos A}{1-sin A}\)-tan A = \(\frac{cos A(1+ sin A)}{(1- sin A)(1+ sin A)}\)- tan A =\(\frac{cos A(1+sin A)}{1+sin^2 A}\)- tan A =\(\frac{cos A(1+ sin A)}{cos ^2 A}\)- tan A =\(\frac{1+ sin A}{cos A}\)- tan A =\(\frac{1}{cos A}\)+\(\frac{sin A}{cos A}\)- tan A = sec A + tan A - tan A  = sec A.

Q3. Prove that : (i)\(\sqrt{\frac{1-cos\theta}{1+cos \theta}}\)= cosec θ - cos θ (ii) \(\sqrt{\frac{1+cos \theta}{1-cos \theta}}\)= sec θ - tan θ

Explanation: (i) L.H.S. =\(\sqrt{\frac{1-cos\theta}{1+cos\theta}×\frac{1-cos\theta}{1-cos\theta}}\) =\(\sqrt{\frac{(1-cos\theta)^2}{1+cos^2\theta}}\) =\(\frac{1-cos\theta}{\sqrt{1-cos^2\theta}}\) =\(\frac{1-cos\theta}{\sqrt{sin^2\theta}}\) =\(\frac{1-cos \theta}{sin \theta}\) =\(\frac{1}{sin \theta}-\frac{cos \theta}{sin \theta}\) = cosec θ – cot θ = R.H.S. Hence Proved. (ii)L.H.S.= \(\sqrt{\frac{1+sin \theta}{1-sin\theta}×\frac{1+sin \theta}{1+sin\theta}}\) =\(\sqrt{\frac{(1+sin\theta)^2}{1-sin^2\theta}}\) =\(\frac{1+sin\theta}{cos\theta}\)=\(\frac{1}{cos\theta}\)+\(\frac{sin\theta}{cos \theta}\) = sec θ + tan θ = R.H.S. Hence Proved

Q4. Prove that : sin 4 θ – cos 4 θ = sin 2 θ – cos 2 θ = 2 sin 2 θ – 1 = 1 – 2 cos 2 θ.

Explanation: Consider, sin 4 θ – cos 4 θ = (sin 2 θ) 2 – (cos 2 θ) 2 = (sin 2 θ – cos 2 θ)(sin 2 θ + cos 2 θ) [∵ (a – b)(a + b) = a 2 – b 2 ] = (sin 2 θ – cos 2 θ) × 1 [∵ sin 2 θ+ cos 2 θ= 1] = sin 2 θ – cos 2 θ = sin 2 θ – (1 – sin 2 θ) [∵cos 2 θ = 1 – sin 2 θ] = sin 2 θ – 1 + sin 2 θ = 2 sin 2 θ – 1 = 2(1 – cos 2 θ) – 1 [∵ sin 2 θ =1 – cos 2 θ] = 2 – 2 cos 2 θ – 1 = 1 – 2 cos 2 θ. Hence Proved.

Q5. The angle of elevation of a cloud from a point 200 metres above a lake is 30° and the angle of depression of its reflection in the lake is 60°. Find the height of the cloud.

Explanation: Let P be the point of observation and C, the position of cloud. CN perpendicular from C on the surface of the lake and C‘ be the reflection of the cloud in the lake so that

Trigonometry_Q5

CN = NC´ = x (say) Then, PM = 200 m ∴   AN = MP = 200 m CA = CN – AN = (x – 200) m C´A = NC´ + AN = (x + 200) m Let PA = y m Then, in right angled Δ  PAC, \(\frac{CA}{PA}\)= tan 30 ⟹ \(\frac{x+200}{y}\)=\(\sqrt{3}\) ⇒ x + 200 = \(\sqrt{3}\)y ⟹y = \(\frac{y+200}{\sqrt{3}}\) ...( ii ) From equations (i) and (ii), \(\frac{x+200}{y}\) =\(\sqrt{3}\)(x-200) ⇒        x +200 = 3( x –200) ⇒        x +200 = 3 x –600 ⇒        2 x = 800 ⇒        x = 400 m ‍ Hence, the height of the cloud = 400 m.

trigonometry class 10 icse important questions

The study of trigonometry in Class 10 ICSE mathematics is essential for understanding the relationships between angles and sides in right triangles. It equips students with tools to solve practical problems and lays the foundation for more advanced mathematics and scientific applications. If you seek additional practice and a deeper comprehension of the topics covered in the chapter, oswal.io offers an extensive array of class 10 Trigonometry important questions and answers to facilitate a more profound understanding of the concepts.

Q1 : What is trigonometry, and why is it important?

Ans: Trigonometry is the study of the relationships between angles and sides in triangles. It's important because it helps us solve real-world problems involving measurements, distances, and angles.

Q2: What are the primary trigonometric ratios?

Ans:  The primary trigonometric ratios are sine (sin), cosine (cos), and tangent (tan).

Q3 : What is the sine of an angle in a right triangle?

Ans: The sine (sin) of an angle is the ratio of the length of the side opposite the angle to the length of the hypotenuse.

Q4 : What is the cosine of an angle in a right triangle?

Ans: The cosine (cos) of an angle is the ratio of the length of the side adjacent to the angle to the length of the hypotenuse.

Q5 : How is the tangent of an angle calculated in a right triangle?

Ans: The tangent (tan) of an angle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.

Chapter Wise  Important Questions for ICSE Board Class 10 Mathematics

trigonometry class 10 icse important questions

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Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8

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Table of Contents

Important questions of Chapter 8 – Introduction to Trigonometry of Class 10 are given here for students who want to score high marks in their board exams 2021-22. Students can also download the trigonometry class 10 questions which are available on our website. They can learn and solve the problems offline by downloading trigonometry class 10 questions. The questions which are provided below are as per the latest CBSE syllabus and are designed according to the NCERT book. The questions are formulated after analyzing the previous year’s questions papers, exam trends and latest sample papers. Solving these questions will help students to get prepared for the final exam. Students can also get important questions for all the chapters of 10th standard Maths. Solve them to get acquainted with the various types of questions to be asked from each chapter of the Maths subject.

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Target Exam ---

In Chapter 8, students will be introduced to the Trigonometry concept, which states the relationship between angles and sides of a triangle. They will come across many trigonometric formulas which will be used to solve numerical problems. The six primary trigonometric ratios are sine, cosine, tangent, secant, cosecant and cotangent. The whole trigonometry concept revolves around these ratios sometimes also called functions. Students can learn more about trigonometry class 10 questions, which are provided with complete explanations. Introduction to Trigonometry Class 10 Maths Chapter 8 Important Questions are solved by our expert teachers so that students can understand the problems quickly. Besides, students can also get additional questions on chapter 8 of class 10 maths for practice at the end.

Introduction-To-Trigonometry-CBSE-Class-10-Maths-Extra-Questions-1

NCERT Solutions for Class 10 Maths – Free Download

FAQs on Introduction to Trigonometry – Class 10 Maths Chapter 8

What is trigonometry in class 10 maths.

Trigonometry is a branch of mathematics that deals with the relationships between angles and sides of triangles.

Why is Trigonometry important in Class 10 Maths?

Trigonometry helps solve real-world problems involving measurements of angles and sides, making it a practical skill.

What concepts are covered in Class 10 Trigonometry?

Class 10 Trigonometry covers topics like trigonometric ratios, Pythagoras theorem, and applications of trigonometry.

What are Trigonometric Ratios?

Trigonometric ratios relate angles to sides of a right triangle. They include sine, cosine, and tangent.

How can Trigonometry be applied in real life?

Trigonometry is used in fields like architecture, engineering, astronomy, and navigation for calculations involving angles and distances.

What is Pythagoras Theorem in Trigonometry?

Pythagoras Theorem states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

How are Trigonometric Ratios calculated?

Trigonometric ratios are calculated by dividing the lengths of specific sides of a right triangle.

What are the primary applications of Trigonometry?

Trigonometry is used for calculating heights, distances, angles of elevation, and depression in real-life scenarios.

How can I understand Trigonometry concepts better?

Practice with examples, diagrams, and real-life scenarios helps in better understanding Trigonometry concepts.

Is Trigonometry covered in other math classes too?

Yes, Trigonometry concepts are introduced in higher math classes and continue to be applied in advanced mathematics.

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ICSE / ISC / CBSE Mathematics Portal for K12 Students

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ICSE Board Problems , Trigonometrical Identities

Class 10: Trigonometry – Board Problems

trigonometry tough questions class 10

Question 14: Evaluate without using trigonometric tables:

\displaystyle 2 (\frac{\tan 35^o}{\cot 55^o})^2 + (\frac{\cot 55^o}{\tan 35^o})^2 - 3 (\frac{\sec 40^o}{\mathrm{cosec} 50^o}) \hspace{1.0cm} [2011]

Question 16: Without using trigonometric tables, evaluate

\displaystyle \sin^2 34^o + \sin^2 56^o + 2 \tan 18^o \tan 72^o - \cot ^2 30^o \hspace{1.0cm} [2014]

Question 17: Prove the identity:

\displaystyle (\sin \theta + \cos \theta) (\tan \theta + \cot \theta) = \sec \theta + \mathrm{cosec} \theta \hspace{1.0cm} [2014]

Therefore LHS = RHS. Hence proved.

\text{Question 19: Evaluate: } \displaystyle \frac{3 \sin 72^o}{\cos 18^o}- \frac{\sec 32^o}{\mathrm{cosec} 58^o} \hspace{1.0cm} [2000]

Question 24: Without using trigonometric tables evaluate :

\displaystyle \frac{\sin 35^o \cos 55^o + \cos 35^o \sin 55^o}{\mathrm{cosec} ^2 10^o - \tan^2 80^o} \hspace{1.0cm} [2010]

Question 25: Without using trigonometric tables evaluate :

\displaystyle = \frac{1}{1} = 1

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13 thoughts on “ Class 10: Trigonometry – Board Problems ”

Add Comment

The qustion no.2 Incorrect..

Thank you for pointing out. It was a typing mistake. I have corrected it.

It’s unable to understand the solution of question no.7

There was a typing mistake in the question… i have added one more line of explanation…. it should be easy to understand now.

I fixed it. Thank you for your contribution.

Technical issues from Q 12 Please sort them out

Let me check

Very helpful.

ques 9 has wrong answer

Good catch… it was a typo. We have corrected it.

(1-tanA)² + (1+tanA)² = 2sec²A LHS = (1-tanA)² + (1+tanA)² LHS = 1-2tanA+tan²A+1+2tanA+tan²A LHS = 2 + 2tan²A LHS = 2(1+tan²A) Here, we use formula (1+tan²A)=sec²A LHS = 2sec²A=RHS

By : MANGESH K REPAL

Sir, there are multiple ways of solving a problem. I have also included your way of solving the problem. Thanks

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CBSE Class 10 Sanskrit Paper Analysis 2024: Student Feedback, Difficulty Level and Expert Review

Cbse board 10th sanskrit paper analysis 2024: students can check the cbse higher secondary class 10 sanskrit paper analysis 2024. complete details on the difficulty level of the question paper, the student’s reaction, and the expert’s review. .

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Importance of CBSE Class 10 Sanskrit 2024 Analysis

Cbse class 10 sanskrit paper format 2024.

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CBSE Class 10 Sanskrit Exam 2024: Student Reaction

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Trigonometric Identities Class 10

Trigonometric identities class 10 includes basic identities of trigonometry. When we recall, an equation is considered identical, if the equations are true for all the values of variables involved. Similarly, the trigonometric equation, which involves trigonometry ratios of all the angles, is called a trigonometric identity if it is true for all values of the angles.

In Mathematics, trigonometry is one of the most important and prominent topics to learn. Trigonometry is basically the study of triangles. The term ‘Trigon’ means triangle and ‘metry’ means measurement. Trigonometric identities class 10 consists of trigonometry ratios such as sine, cosine and tangent in its equations. Even, trigonometry identities class 10 formulas are based on these ratios. These identities are used to solve various trigonometry problems.

By considering a right-angled triangle, trigonometry identities class 10 lists could be figured out. The trigonometric identities or equations are formed using trigonometry ratios for all the angles. Using trigonometry identities, we can express each trigonometric ratios in terms of other trigonometric ratios, and if any of the trigonometry ratios value is known to us, then we can find the values of other trigonometric ratios. We can also solve trigonometric identities class 10 questions, using these identities as well.

Trigonometric Identities for Class 10

Trigonometric identities are the equations that include the trigonometric functions such as sine, cosine, tangent, etc., and are true for all values of angle θ. Here, θ is the reference angle taken for a right-angled triangle. In class 10th, there are basically three trigonometric identities, which we learn in the trigonometry chapter. They are:

  • Cos 2 θ + Sin 2 θ = 1
  • 1 + Tan 2 θ = Sec 2  θ
  • 1 + Cot 2  θ = Cosec 2  θ

Here, we will prove one trigonometric identity and will use it to prove the other two. Take an example of a right-angled triangle ΔABC.

Trigonometric Identities Class 10 Proof

Proof of Trigonometric Identities Class 10

In a right-angled triangle, by the Pythagorean theorem , we know,

(Perpendicular) 2 + (Base) 2 = (Hypotenuse) 2

Therefore, in ΔABC, we have;

AB 2 + BC 2 = AC 2  ….. (1)

Dividing equation (1) by AC 2 , we get,

Cos 2 θ + Sin 2 θ = 1 …..(2)

If θ = 0, then,

  • Cos 2 0 + Sin 2 0 = 1
  • 1 2 + 0 2 = 1

And if we put θ = 90,then

  • Cos 2 90 + Sin 2 90 = 1
  • 0 2 + 1 2 = 1

For all angles, 0°≤ θ ≤ 90°, equation (2) is satisfied. Hence, equation (2) is a trigonometric identity.

Again, divide equation (1) by AB 2 , we get

1 + Tan 2 θ = Sec 2  θ …..(3)

If θ = 0, then,

  • 1 + tan 2 0 = sec 2 0
  • 1 + 0 2 = 1 2
  • 1 + tan 2 90 = sec 2 90

As you can see, the values of both sides are equal. Therefore, it proves that for all the values between 0 0 and 90 0 , the equation (3) is satisfied. So, it is also a trigonometric identity.

Let’s see what we get if we divide equation (1) by BC 2 , we get,

Cot 2  θ + 1 = Cosec 2 θ ….(4)

Now let us prove this identity as well.

If θ = 0, then equation (4) can be written as;

  • Cot 2 0 + 1 = Cosec 2 0

Both sides are equal.

And if θ = 90,then equation (4) can be written as;

  • Cot 2 90 + 1 = Cosec 2 90
  • 0 2 + 1 = 1 2

Thus, proved that equation (4) is a trigonometric identity.

Video Lesson on Trigonometry

trigonometry tough questions class 10

More Formulas for Class 10

  • Geometry Formulas For Class 10
  • Algebra formulas for Class 10
  • Maths Formulas For Class 10
  • Mensuration Formulas Class 10
  • Trigonometry Formulas For Class 10

Practice Questions From Class 10 Trigonometry Identities

  • Prove √(sec θ – 1)/(sec θ + 1) = cosec θ – cot θ
  • Prove (tan θ + sec θ – 1)/(tan θ – sec θ + 1) = (1 + sin θ)/cos θ
  • Prove sec θ√(1 – sin 2  θ) = 1
  • Given, √3 tan θ = 3 sin θ. Prove sin 2  θ – cos 2  θ = 1/3
  • Evaluate cos 2  θ tan 2  θ + tan 2  θ sin 2  θ in terms of tan θ.

Practice Now:  Important Questions Class 10 Maths Chapter 8 Introduction Trigonometry

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    Easy Normal Difficult Trigonometry Problems - sin, cos, tan, cot: Very Difficult Problems with Solutions Problem 1 If \displaystyle x+y+z=\pi x+y +z = π prove the trigonometric identity

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    Question. 1 : In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C Solution: In a given triangle ABC, right-angled at B = ∠B = 90° Given: AB = 24 cm and BC = 7 cm That means, AC = Hypotenuse According to the Pythagoras Theorem,

  3. Important Questions for Class 10 Maths Chapter 8 Introduction to

    Introduction to Trigonometry Class 10 Important Questions Very Short Answer (1 Mark) Question 1. If tan θ + cot θ = 5, find the value of tan2θ + cotθ. (2012) Solution: tan θ + cot θ = 5 … [Given tan 2 θ + cot 2 θ + 2 tan θ cot θ = 25 … [Squaring both sides tan 2 θ + cot 2 θ + 2 = 25 ∴ tan 2 θ + cot 2 θ = 23 Question 2.

  4. Extra Questions for Class 10 Maths Trigonometry

    NCERT Solutions NCERT Class 10 Maths Trigonometry Extra Questions Chapter 8 of CBSE NCERT Class 10 Math covers Trigonometry. Concepts covered in Chapter 8 include trigonometric ratios, trigonometric ratios of complementary angles, trigonometric identities.

  5. Important Questions Class 10 Maths Chapter 9 Applications of Trigonometry

    Students who are preparing CBSE 2022-2023 Maths exam are advised to practice these important questions of Some Applications of Trigonometry For Class 10. Solving these questions will help students to score high marks in the questions asked from this chapter.

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    Trigonometry Important Questions for Class 10 Maths Given below are the Class 10 Maths trigonometry extra questions.This includes both important and tough questions a. Multiple Choice Questions b. Short Answer type c. Long answer type d. Fill in the blanks e. True and false Multiple Choice Questions Question 1

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    INTRODUCTION TO TRIGNOMETRY HOTS 1. Prove that 2. If tanA=n tanB and sin A =m sin B, prove that n 2 - 1 3. Prove that following identity, where the angle involved is acute andgle for whhc the expression are expressed using identity cosec 2 A= 1+cot 2 A 4. If x sin 3 |+ y cos 3 |=sin|cos| and xsin|=ycos|, prove x 2 +y 2 =1. 5.

  10. CBSE Class 10 Maths Chapter 8 Introduction To Trigonometry ...

    CBSE Class 10 Maths Chapter 8 Important Questions revolve around the concept of trigonometric equations at its base. The Class 10 Maths Ch 8 Important Questions by Vedantu come with all the solutions that are drafted in a way, to provide you with a maximum understanding of the subject and figure out the different aspects of trigonometry. The important questions include questions from all the ...

  11. Class 10 Maths Chapter 8 Introduction to Trigonometry MCQs

    Class 10 Maths MCQs for Chapter 8 (Introduction to trigonometry) are given here with answers and detailed explanations. All these multiple-choice questions are available online as per the CBSE syllabus (2022 - 2023) and NCERT guidelines. These MCQs for Class 10 Maths Chapter 8 are prepared as per the latest exam pattern.

  12. Introduction to Trigonometry Class 10 Extra Questions Maths Chapter 8

    Question 1. From the given figure, find the value of x: Answer: In the given fig., only one side is known which is hypotenuse and side to be evaluated is BC which is perpendicular with reference to given angle ZA = 30°. ∴ sin 30° = ⇒ x = 15 sin 30° x = 15 × = 7 cm Question 2. If tan A = cot B, prove that A + B = 90°. Answer: ∵ tan A = cot B

  13. Trigonometry Class 10

    To study the answers of the NCERT Questions, click on an exercise or topic below. The chapter is updated according to thenew NCERT, for 2023-2024 Board Exams.Get NCERT Solutions with videos of all questions and examples of Chapter 8 Class 10 Trigonometry. Videos of all questions are made with step-by-step explanations.

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    Solution: ∠BAC = 180° - 90° - 60o = 30° sin 30° = BC AC 1 2 = BC 15 2BC = 15 BC = 15 2 m Question 2. In the given figure, a tower AB is 20 m high and BC, its shadow on the ground, is 20 3-√ m long. Find the Sun's altitude. (2015OD) Solution: AB = 20 m, BC = 20 3-√ m, θ = ? In ∆ABC, Question 3.

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  19. NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

    NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry is helpful for the students as it aids in understanding the concepts as well as in scoring well in CBSE Class 10 board examination. The NCERT Solutions are designed and reviewed by subject experts and cover all the questions from the textbook.

  20. Class 10 Maths Chapter 8 Introduction to Trigonometry Extra Questions

    Important questions of Chapter 8 - Introduction to Trigonometry of Class 10 are given here for students who want to score high marks in their board exams 2021-22. Students can also download the trigonometry class 10 questions which are available on our website. They can learn and solve the problems offline by downloading trigonometry class 10 ...

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    Class 10: Trigonometry - Board Problems Date: January 15, 2018 ICSE CBSE ISC Board Mathematics Portal for Students 13 Comments Answer: Answer: OR Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Answer: Question 14: Evaluate without using trigonometric tables: Answer: Answer:

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    Applications of Trigonometry Class 10 Extra Questions HOTS. Question 1. A spherical balloon of radius r subtends an angle 9 at the eye of an observer. If the angle of elevation of its centre is $, find the height of the centre of the balloon. Answer: Let O be the centre of the balloon of radius .

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  24. Trigonometric Identities Class 10

    1 + Cot 2 θ = Cosec 2 θ Here, we will prove one trigonometric identity and will use it to prove the other two. Take an example of a right-angled triangle ΔABC. Proof of Trigonometric Identities Class 10 In a right-angled triangle, by the Pythagorean theorem, we know, (Perpendicular) 2 + (Base) 2 = (Hypotenuse) 2 Therefore, in ΔABC, we have;