NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

Explore the NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry (English Medium), with detailed explanations for Exercise 8.4. This resource is designed to simplify complex Introduction to Trigonometry, helping Class 10 Maths students grasp the subject effortlessly. For any queries regarding NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4, feel free to leave a comment, and we’ll respond as soon as possible. || Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.4 || NCERT Solutions for Class 10 Maths Chapter 8: Introduction to Trigonometry (English Medium)

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4
Exercise 8.4

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

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1. Express the trigonometric ratios $ \sin A, \sec A $ and $ \tan A $ in terms of $ \cot A $.

Answer

We know that,
$ \operatorname{cosec}^{2} A=1+\cot ^{2} A $
$ \frac{1}{\operatorname{cosec}^{2} A}=\frac{1}{1+\cot ^{2} A} $
$ \sin ^{2} A=\frac{1}{1+\cot ^{2} A} $
$ \sin A= \pm \frac{1}{\sqrt{1+\cot ^{2} A}} $
$ \sqrt{1+\cot ^{2} A} $ will always be positive as we are adding two positive quantities.
Therefore, $ \sin A=\frac{1}{\sqrt{1+\cot ^{2} A}} $
We know that, $ \tan A=\frac{\sin A}{\cos A} $
However, $ \cot A=\frac{\cos A}{\sin A} $
Therefore, $ \tan A=\frac{1}{\cot A} $
Also, $ \sec ^{2} A=1+\tan ^{2} A $
$ =1+\frac{1}{\cot ^{2} A} $
$ =\frac{\cot ^{2} A+1}{\cot ^{2} A} $
$ \sec A=\frac{\sqrt{\cot ^{2} A+1}}{\cot A} $

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

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2. Write all the other trigonometric ratios of $ \angle \mathrm{A} $ in terms of $ \sec A $.

Answer

We know that,
$\cos A=\frac{1}{\sec A}$
Also, $\sin ^{2} A+\cos ^{2} A=1$
$\sin ^{2} A=1-\cos ^{2} A$
$\sin A=\sqrt{1-\left(-\frac{1}{\sec ^{2} A}\right)^{2}}$
$=\sqrt{\frac{\sec ^{2} A-1}{\sec ^{2} A}}=\frac{\sqrt{\sec ^{2} A-1}}{\sec A}$
$=\sqrt{\frac{\sec ^{2} A}{\sec ^{2} A}}=\frac{\sqrt{\sec ^{2} A-1}}{\sec A}$
$\tan ^{2} A=\sec ^{2} A-1$
$\tan A=\sqrt{\sec ^{2} A-1}$
$\cot A=\frac{\cos A}{\sin A}=\frac{\sec A}{\sec A}$
$=\frac{1}{\sqrt{\sec ^{2} A-1}}$
$\operatorname{cosec} A=\frac{1}{\sin A}=\frac{\sec A}{\sqrt{\sec ^{2} A-1}}$

(i) Evaluate
$ \frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}} $

Answer

$\frac{\sin ^{2} 63^{\circ}+\sin ^{2} 27^{\circ}}{\cos ^{2} 17^{\circ}+\cos ^{2} 73^{\circ}}$
$=\frac{\left[\sin \left(90^{\circ}-27^{\circ}\right)\right]^{2}+\sin ^{2} 27^{\circ}}{\left[\cos \left(90^{\circ}-73^{\circ}\right)\right]^{2}+\cos ^{2} 73^{\circ}}$
$=\frac{\left(\cos 27^{\circ}\right]^{2}+\sin ^{2} 27^{\circ}}{\left[\sin 7 3^{\circ}\right]^{2}+\cos ^{2} 73^{\circ}}$
$=\frac{\cos ^{2} 27^{\circ}+\sin ^{2} 27^{\circ}}{\sin ^{2} 73^{\circ}+\cos ^{2} 73^{\circ}}$
$=\frac{1}{1}\left(\text { As} \sin ^{2} \mathrm{~A}+\cos ^{2} \mathrm{~A}=1\right)$
$=1$

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

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(ii) Evaluate
$ \sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ} $

Answer

$ \sin 25^{\circ} \cos 65^{\circ}+\cos 25^{\circ} \sin 65^{\circ}$
$=\left(\sin 25^{\circ}\right)\left\{\cos \left(90^{\circ}-25^{\circ}\right)\right\}+\cos 25^{\circ}\left\{\sin \left(90^{\circ}-25^{\circ}\right)\right\}$
$=\left(\sin 25^{\circ}\right)\left(\sin 25^{\circ}\right)+\left(\cos 25^{\circ}\right)\left(\cos 25^{\circ}\right)$
$=\sin 25^{\circ}+\cos 25^{\circ}$
$=1\left(\text { As } \sin ^{2} A+\cos ^{2} A=1\right)$

(i) Choose the correct option. Justify your choice.
$ 9 \sec ^{2} A-9 \tan ^{2} A= $
(A) 1 (B) 9 (C) 8 (D) 0

Answer

$ 9 \sec ^{2} \mathrm{~A}-9 \tan ^{2} \mathrm{~A}$
$=9\left(\sec ^{2} \mathrm{~A}-\tan ^{2} \mathrm{~A}\right)$
$=9(1)\left[\mathrm{As} \sec ^{2} \mathrm{~A}-\tan ^{2}\mathrm{~A}=1\right]$
$=9$
Hence, alternatives $ (B) $ is correct.

(ii) Choose the correct option. Justify your choice.
$ (1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)= $
(A) 0 (B) 1 (C) 2 (D) -1

Answer

$ (1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)$
$= \left(1+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(1+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right)$
$= \left(\frac{\cos \theta+\sin \theta+1}{\cos \theta}\right)\left(\frac{\sin \theta+\cos \theta-1}{\sin \theta}\right)$
$= \frac{(\sin \theta+\cos \theta)^{2}-(1)^{2}}{\sin \theta \cos \theta}$
$= \frac{\sin ^{2} \theta+\cos ^{2} \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}$
$= \frac{2 \sin \theta+\cos \theta+2 \sin \theta \cos \theta-1}{\sin \theta \cos \theta}=2$
$= \frac{2 \sin \theta \cos \theta}{\sin \theta \cos \theta}=2$
Hence, alternative (C) is incorrect.

(iii) Choose the correct option. Justify your choice.
$ (\sec A+\tan A)(1-\sin A)= $
(A) $ \sec A $ (B) $ \sin A $ (C) $ \operatorname{cosec} A $ (D) $ \cos A $

Answer

$ (\sec A+\tan \mathrm{A})(1-\sin \mathrm{A})$
$=\left(\frac{1}{\cos A}+\frac{\sin A}{\cos A}\right)(1-\sin A)$
$=\left(\frac{1+\sin A}{\cos A}\right)(1-\sin A)$
$=\frac{1-\sin ^{2} A}{\cos A}=\frac{\cos ^{2} A}{\cos A}$
$=\cos A$
Hence, alternative (D) is correct.

(iv) Choose the correct option. Justify your choice.
$ \frac{1+\tan ^{2} A}{1+\cot ^{2} A}= $
(A) $ \sec ^{2} A $ (B) $-1$ (C) $ \cot ^{2} \mathrm{~A} $ (D) $ \tan ^{2} \mathrm{~A} $

Answer

$ \frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}} $
$ =\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}} $
$ =\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A $
Hence, alternative (D) is correct.

(i) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ (\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta} $

Answer

$ (\operatorname{cosec} \theta-\cot \theta)^{2}=\frac{1-\cos \theta}{1+\cos \theta} $
L.H.S. $ =(\operatorname{cosec} \theta-\cot \theta)^{2} $
$ =\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2} $
$ =\frac{(1-\cos \theta)^{2}}{(\sin \theta)^{2}}-\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta} $
$ =\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}-\frac{(1-\cos \theta)^{2}}{(1-\cos \theta)(1+\cos \theta)}=\frac{1-\cos \theta}{1+\cos \theta} $
$ = $ R.H.S.

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

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(ii) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A $

Answer

$ \frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}=2 \sec A $
$\text { L.H.S. }=\frac{\cos A}{1+\sin A}+\frac{1+\sin A}{\cos A}$
$=\frac{\cos ^{2} A+(1+\sin A)^{2}}{(1+\sin A)(\cos A)^{2}}$
$=\frac{\cos ^{2} A+1+\sin ^{2} A+2 \sin A}{(1+\sin A)(\cos A)}$
$=\frac{\sin ^{2} A+\cos ^{2} A+1+2 \sin A}{(1+\sin A)(\cos A)}$
$=\frac{1+1+2 \sin A}{(1+\sin A)(\cos A)}=\frac{2+2 \sin A}{(1+\sin A)(\cos A)}$
$=\frac{2(1+\sin A)}{(1+\sin A)(\cos A)}=\frac{2}{\cos A}=2 \sec A$
$=R . H . S \text {. }$

(iii) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta $

Answer

$ \frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\sec \theta \operatorname{cosec} \theta $
L.H.S. $ =\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta} $
$ =\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}} +\frac{\frac{\cos \theta}{\sin \theta}}{1 - \frac{\sin \theta}{\cos \theta}} $
$ =\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}+\frac{\cos \theta}{\cos \theta}} $
$=\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)}$
$ =\frac{1}{(\sin \theta-\cos \theta)}\left[\frac{\sin ^{2} \theta}{\cos \theta} - \frac{\cos^2 \theta}{\sin \theta} \right] $
$=\left(\frac{1}{\sin \theta-\cos \theta}\right)\left[\frac{\sin ^{3} \theta-\cos ^{3} \theta}{\sin \theta \cos \theta}\right]$
$=\left(\frac{1}{\sin \theta-\cos \theta}\right)\left[\frac{(\sin \theta-\cos \theta)\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \cos \theta\right)}{\sin \theta \cos \theta}\right]$
$=\frac{(1+\sin \theta \cos \theta)}{(\sin \theta \cos \theta)}$
$ =\sec \theta \operatorname{cosec} \theta $
$=R . H . S$

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

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(iv) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ \frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A} $

Answer

$ \frac{1+\sec A}{\sec A}=\frac{\sin ^{2} A}{1-\cos A} $
L. H.S. $ =\frac{1+\sec A}{\sec A}=\frac{1+\frac{1}{\cos A}}{\frac{1}{\cos A}} $
$=\frac{\cos A+1}{\cos A}=(\cos A+1)$
$=\frac{(1-\cos A)(1+\cos A)}{(1-\cos A)}$
$=\frac{1-\cos ^{2} A}{1-\cos A}=\frac{\sin ^{2} A}{1-\cos A}$
$ = $ R.H.S.

(v) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ \frac{\cos A - \sin A+1}{\cos A + \sin A-1} $ $= \operatorname{cosec} A + \cot A$, using the identity $\operatorname{cosec}^{2} A= 1 + cot^{2} A$

Answer

$ \frac{\cos A-\sin A+1}{\cos A+\sin A-1}=\operatorname{cosec} A+\cot A $
Using the identity $ \operatorname{cosec}^{2} A=1+\cot ^{2} A $
$\text { L.H.S. }=\frac{\cos A-\sin A+1}{\cos A+\sin A-1}$
$=\frac{\frac{\cos A \sin A}{\sin A}+\frac{1}{\sin A}}{\frac{\sin A}{\sin A}+\frac{\sin A}{\sin A}+\frac{1}{\sin A}}$
$=\frac{\cot A-1+\operatorname{cosec} A}{\cot A+1-\operatorname{cosec} A}$
$=\frac{\{(\cot A)-(1-\operatorname{cosec} A)\}\{(\cot A)-(1-\operatorname{cosec} A)\}}{\{(\cot A)+(1-\operatorname{cosec} A)\}\{(\cot A)-(1-\operatorname{cosec} A)\}}$
$=\frac{(\cot A-1+\operatorname{cosec} A)^{2}}{(\cot A)^{2}-(1-\operatorname{cosec} A)^{2}}$
$=\frac{\cot ^{2} A+1+\operatorname{cosec}^{2} A-2 \cot A-2 \operatorname{cosec} A+2 \cot A \operatorname{cosec} A}{\cot ^{2} A-\left(1+\operatorname{cosec}^{2} A-2 \operatorname{cosec} A\right)}$
$=\frac{2 \operatorname{cosec}^{2} A+2 \operatorname{cosec} A-2 \cot A-2 \operatorname{cosec} A}{\cot ^{2} A-1-\operatorname{cosec}^{2} A+2 \operatorname{cosec} A}$
$=\frac{2 \operatorname{cosec} A(\operatorname{cosec} A+\cot A)-2(\cot A+\operatorname{cosec} A)}{\cot ^{2} A-\operatorname{cosec}^{2} A-1+2 \operatorname{cosec} A}$
$=\frac{(\operatorname{cosec} A+\cot A)(2 \operatorname{cosec} A-2)}{-1-1+2 \operatorname{cosec} A}$
$=\frac{(\operatorname{cosec} A+\cot A)(2 \operatorname{cosec} A-2)}{(2 \operatorname{cosec} A-2)}$
$=\operatorname{cosec} A+\cot A$
$=R . H . S$

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

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(vi) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ \sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A $

Answer

$ \sqrt{\frac{1+\sin A}{1-\sin A}}=\sec A+\tan A $
$\text { L.H.S. }=\sqrt{\frac{1+\sin A}{1-\sin A}}$
$\quad=\sqrt{\frac{(1+\sin A)(1+\sin A)}{(1-\sin A)(1+\sin A)}}$
$=\frac{(1+\sin A)}{\sqrt{1-\sin ^{2} A}}=\frac{1+\sin A}{\sqrt{\cos ^{2} A}}$
$=\frac{1+\sin A}{\cos A}=\sec A+\tan A$
$=\text { R.H.S. }$

(vii) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}=\tan \theta $

Answer

$ \frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos \theta \cos \theta}=\tan \theta $
$\text { L.H.S. } =\frac{\sin \theta-2 \sin ^{3} \theta}{2 \cos ^{3} \theta-\cos \theta}$
$ =\frac{\sin \theta\left(1-2 \sin ^{2} \theta\right)}{\cos \theta\left(2 \cos ^{2} \theta-1\right)}$
$=\frac{\sin \theta \times\left(1-2 \sin ^{2} \theta\right)}{\cos \theta \times\left\{2\left(1-\sin ^{2} \theta\right)-1\right\}}$
$=\frac{\sin \theta \times\left(1-2 \sin ^{2} \theta\right)}{\cos \theta \times\left(1-2 \sin ^{2} \theta\right)}$
$=\tan \theta=\text { R.H.S }$

(viii) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ (\sin A+\operatorname{cosec} A)^{2}+(\cos A+\sec A)^{2}=7+\tan ^{2} A+\cot ^{2} A $

Answer

$(\sin A+\operatorname{cosec} A)^{2}+(\cos A+\sec A)^{2}=7+\tan ^{2} A+\cot ^{2} A$
$\text { L.H.S }=(\sin A+\operatorname{cosec} A)^{2}+(\cos A+\sec A)^{2}$
$=\sin ^{2} A+\operatorname{cosec}^{2} A+2 \sin A \operatorname{cosec} A+\cos ^{2} A+\sec ^{2} A+2 \cos A \sec A$
$=\left(\sin ^{2} A+\cos ^{2} A\right)+\left(\operatorname{cosec}^{2} A+\sec ^{2} A\right)+2 \sin A\left(\frac{1}{\sin A}\right)+2 \cos A\left(\frac{1}{\cos A}\right)$
$=(1)+\left(1+\cot ^{2} A+1+\tan ^{2} A\right)+(2)+(2)$
$=7+\tan ^{2} A+\cot ^{2} A$
$=\text { R.H.S }$

(ix) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$ (\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A} $

Answer

$(\operatorname{cosec} A-\sin A)(\sec A-\cos A)=\frac{1}{\tan A+\cot A}$
$\text { L.H.S }=(\operatorname{cosec} A-\sin A)(\sec A-\cos A)$
$=\left(\frac{1}{\sin A}-\sin A\right)\left(\frac{1}{\cos A}-\cos A\right)$
$ =\left(\frac{1-\sin ^{2} A}{\sin A}\right)\left(\frac{1-\cos ^{2} A}{\cos A}\right) $
$=\frac{\left(\cos ^{2} A\right)\left(\sin ^{2} A\right)}{\sin A \cos A}$
$=\sin A \cos A$
$\text { R.H.S }=\frac{1}{\tan A+\cot A}$
$ =\frac{1}{\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}} = \frac{1}{\frac{\sin ^{2} A+\cos ^{2} A}{\sin A \cos A}} $
$=\frac{\sin A \cos A}{\sin ^{2} A+\cos ^{2} A}=\sin A \cos A$
Hence, L.H.S = R.H.S

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

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(x) Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A $

Answer

$\left(\frac{1+\tan ^{2} A}{1+\cot ^{2} A}\right)=\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\tan ^{2} A$
$\frac{1+\tan ^{2} A}{1+\cot ^{2} A}=\frac{1+\frac{\sin ^{2} A}{\cos ^{2} A}}{1+\frac{\cos ^{2} A}{\sin ^{2} A}}=\frac{\frac{\cos ^{2} A+\sin ^{2} A}{\cos ^{2} A}}{\frac{\cos ^{2} A}{\sin ^{2} A}}$
$=\frac{\frac{1}{\cos ^{2} A}}{\frac{1}{\sin ^{2} A}}=\frac{\sin ^{2} A}{\cos ^{2} A}$
$=\tan ^{2} A$
$\left(\frac{1-\tan A}{1-\cot A}\right)^{2}=\frac{1+\tan ^{2} A-2 \tan A}{1+\cot ^{2} A-2 \cot A}$
$=\frac{\sec ^{2} A-2 \tan A}{\operatorname{cosec}^{2} A-2 \cot A}$
$=\frac{\frac{1}{\cos ^{2} A}-\frac{2 \sin A}{\cos A}}{\frac{1}{\sin ^{2} A}-\frac{2 \cos A}{\sin A}}=\frac{\frac{1-2 \sin A \cos A}{\cos ^{2} A}}{\frac{1-2 \sin ^{2} A}{\sin ^{2} A}}$
$=\frac{\sin ^{2} A}{\cos ^{2} A}=\tan ^{2} A$

NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

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NCERT Solutions for Class 10 Maths Chapter 8 Exercise 8.4 || CBSE class 10 maths chapter 8 Ex 8.4

Download the Math Ninja App Now