Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 2

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Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 1

Question 11. Prove $\tan^{-1}(\frac{\sqrt{(1+ x )}-\sqrt{(1-x)}}{\sqrt{(1+ x) }+\sqrt{(1-x)}})=\frac{\pi}{4}-\frac{1}{2} \cos^{-1}x,-\frac{1}{\sqrt2}\le x \le1$

Solution:

Put $x=\cos2\theta$ so that, $\theta= \frac{1}{2} \cos^{-1} x.$

Then, we have :

LHS = $\tan^{-1}(\frac{\sqrt{(1+ x) }-\sqrt{(1-x)}}{\sqrt{(1+ x )}+\sqrt{(1-x)}})$

= $\tan^{-1}(\frac{\sqrt{(1+ \cos2 \theta) }-\sqrt{(1-\cos2 \theta)}}{\sqrt{(1+ \cos2 \theta) }+\sqrt{(1-\cos2 \theta)}})$

= $\tan^{-1}(\frac{\sqrt{(2 \cos^{2} \theta) }-\sqrt{(2 \sin^{2}\theta)}}{\sqrt{(2 \cos^{2}\theta) }+\sqrt{(2 \sin^{2} \theta)}})$

= $\tan^{-1}(\frac{\sqrt{2 }\cos \theta -{\sqrt{2 }\cos\theta}}{{\sqrt{2 }\cos\theta }+{\sqrt{2 }\cos \theta}})$

= $\tan^{-1}(\frac{\cos \theta - \sin \theta}{\cos \theta + \sin \theta})=\tan^{-1}(\frac{1 - \tan \theta}{1 + \tan \theta})$

$\tan^{-1} 1- \tan^{-1}(\tan \theta)$ – $[\tan^{-1}(\frac{x-y}{1+xy})]=\tan^{-1}x+\tan^{-1}y$

$=\frac{\pi}{4} -\theta = \frac{\pi}{4}-\frac{1}{2} \cos^{-1} x$

L.H.S = R.H.S

Hence Proved

Question 12. Prove $\frac{9\pi}{8}- \frac{9}{4} \sin^{-1} \frac{1}{3}=\frac{9}{4} \sin^{-1} \frac{2\sqrt2}{3}$

Solution:

L.H.S. = $\frac{9\pi}{8}- \frac{9}{4} \sin^{-1} \frac{1}{3}$

= $\frac {9}{4}(\frac{\pi}{2}- \sin^{-1} \frac{1}{3})$

Using $\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$

= $\frac {9}{4}(\cos^{-1} \frac{1}{3})$ -(1)

Now, let $\cos^{-1}\frac{1}{3}=x$ Then, $\cos x =\frac{1}{3} \implies \sin x=\sqrt{1-(\frac{1}{3})^{2}}=\frac{2\sqrt{2}}{3}$

$\therefore x=\sin^{-1} \frac{2\sqrt2}{3}$

Using equation(1), we get,

= $\frac{9}{4}\sin^{-1} \frac{2\sqrt2}{3}$

L.H.S = R.H.S

Hence Proved

Question 13. Solve $2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)$

Solution:

$2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)$

= $\tan^{-1}(\frac{2 \cos x}{1- \cos^{2}x})=\tan^{-1} (2\cosec x)$ – $[2 \tan^{-1} x=\tan^{-1} \frac{2x}{1-x^{2}}]$

= $\frac{2\cos x} {1-cos^{2}x}=2\cosec x$

= $\frac{ 2 \cos x}{\sin^{2}}= \frac{2}{\sin x}$

= cos x/sin x

= cot x =1

Therefore, x = π/4

Question 14. Solve $\tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x,(x>0)$

Solution:

$\tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x$

Let x = tanθ

$\tan^{-1}(\frac{1-tan\theta}{1+tan\theta})=\frac{1}{2} \tan^{-1} tan\theta$

$\tan^{-1}(\frac{tan\frac{\pi}{4}-tan\theta}{tan\frac{\pi}{4}+tan\theta})=\frac{1}{2} \theta$

$\tan^{-1}tan(\frac{\pi}{4}-\theta)=\frac{\theta }{2}$

π/4 – θ = θ/2

θ = π/6

So, x = tan(π/6) = 1/√3

Question 15. Solve $\sin(\tan^{-1}x),|x|<1$ is equal to

(A) $\frac{x}{\sqrt{(1-x^{2})}}$ (B) $\frac{1}{\sqrt{(1-x^{2})}}$ (C) $\frac{1}{\sqrt{(1+x^{2})}}$ (D) $\frac{x}{\sqrt{(1+x^{2})}}$

Solution:

Let tan y = x, $\sin y = \frac{x}{\sqrt{(1+x^{2})}}$

Let $\tan^{-1} x=y$ Then,

$\therefore y=\sin^{-1}(\frac{x}{\sqrt{(1+x^{2})}}) \implies \tan^{-1}x=\sin^{-1}\frac{x}{\sqrt{(1+x^{2})}}$

$\therefore \sin(\tan^{-1}x)=sin(sin^{-1}\frac{x}{\sqrt{(1+x^{2})}})=\frac{x}{\sqrt{(1+x^{2})}}$

So, the correct answer is D.

Question 16. Solve $\sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2}$ , then x is equal to

(A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2

Solution:

$\sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2}$

$\implies -2 \sin^{-1}x=\frac{\pi}{2} - \sin^{-1}(1-x)$

$\implies -2 \sin^{-1}x=\cos^{-1}(1-x)$ -(1)

Let $\sin^{-1} x =\theta \to \sin \theta=x$

$\cos \theta= \sqrt{1-x^{2}}$

$\therefore \theta= \cos^{-1}(\sqrt{1-x^{2}})$

$\therefore \sin^{-1} x=cos^{-1}(\sqrt{1-x^{2}})$

Therefore, from equation(1), we have

$-2cos^{-1}(\sqrt{1-x^{2}})=\cos^{-1}(1-x)$

Put x = siny then, we have:

$-2\cos^{-1}(\sqrt{1-\sin^{2} y})=\cos^{-1}(1-\sin y)$

$-2 \cos^{-1}(\cos y)=\cos^{-1}(1-\sin y)$

$-2 y=\cos^{-1}(1-\sin y)$

$1- \sin y=\cos (-2y)=\cos 2y$

$1- \sin y= 1- 2 \sin^{2} y$

$2\sin^{2} y- \sin y=0$

$\sin y(2 \sin y-1)=0$

sin y = 0 or 1/2

x = 0 or x = 1/2

But, when x = 1/2 it can be observed that:

L.H.S. = $\sin^{-1}(1-\frac{1}{2}) -2\sin^{-1} \frac{1}{2}$

= $\sin^{-1} (\frac{1}{2})-2\sin^{-1} \frac{1}{2}$

= $-\sin^{-1} \frac{1}{2}$

= $- \frac{\pi}{6} \ne\frac{\pi}{2}\ne \space R.H.S.$

x = 1/2 is not the solution of given equation.

Thus, x = 0

Hence, the correct answer is C

Question 17. Solve $\tan ^{-1}(\frac{x}{y})-\tan ^{-1}(\frac{x-y}{x+y})$ is equal to

(A) π/2 (B) π/3 (C) π/4 (D) -3π/4

Solution

$\tan^{-1} (\frac{x}{y})-\tan^{-1} \frac{x-y}{x+y}$

$\tan^{-1}[\frac{\frac{x}{y}-\frac{x-y}{x+y}}{1+\frac{x}{y} \times\frac{x-y}{x+y}}]$ – $[\tan^{-1}x+\tan^{-1}y=[\tan^{-1}(\frac{x-y}{1+xy})]]$

$\tan^{-1}[\frac {\frac {x(x+y)-y(x-y)} {y(x+y)} } {\frac {y(x+y)+x(x-y)} {y(x+y)}}]$

${\tan}^{-1}[{\frac {x^2+xy-xy+y^2} {xy+y^2+x^2-xy}}]$

${\tan}^{-1}[\frac {x^2+y^2} {x^2+y^2}]=tan^{-1}1=\frac {\pi} {4}$

Hence, the correct answer is C

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Last Updated : 05 Apr, 2021

Class 12 NCERT Solutions- Mathematics Part I - Chapter 2 Inverse Trigonometric Functions - Miscellaneous Exercise on Chapter 2 | Set 1

Class 12 NCERT Solutions- Mathematics Part I - Chapter 3 Matrices - Exercise 3.1

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Applications of Derivatives

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 3 - Matrices

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Applications of Derivatives

Class 12 NCERT Solutions- Mathematics Part I – Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 2

Chapter 2 Inverse Trigonometric Functions – Miscellaneous Exercise on Chapter 2 | Set 1

Question 11. Prove

Question 12. Prove

Question 13. Solve

Question 14. Solve

Question 15. Solve is equal to

(A) (B) (C) (D)

Question 16. Solve , then x is equal to

(A) 0, 1/2 (B) 1, 1/2 (C) 0 (D) 1/2

Question 17. Solve is equal to

(A) π​/2 (B) π​/3 (C) π​/4 (D) -3π​/4

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Question 11. Prove $\tan^{-1}(\frac{\sqrt{(1+ x )}-\sqrt{(1-x)}}{\sqrt{(1+ x) }+\sqrt{(1-x)}})=\frac{\pi}{4}-\frac{1}{2} \cos^{-1}x,-\frac{1}{\sqrt2}\le x \le1$

Question 12. Prove $\frac{9\pi}{8}- \frac{9}{4} \sin^{-1} \frac{1}{3}=\frac{9}{4} \sin^{-1} \frac{2\sqrt2}{3}$

Question 13. Solve $2\tan^{-1}(\cos x)=\tan^{-1}(2\cosec x)$

Question 14. Solve $\tan^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan^{-1} x,(x>0)$

Question 15. Solve $\sin(\tan^{-1}x),|x|<1$ is equal to

(A) $\frac{x}{\sqrt{(1-x^{2})}}$ (B) $\frac{1}{\sqrt{(1-x^{2})}}$ (C) $\frac{1}{\sqrt{(1+x^{2})}}$ (D) $\frac{x}{\sqrt{(1+x^{2})}}$

Question 16. Solve $\sin^{-1}(1- x)-2 \sin^{-1}x=\frac{\pi}{2}$ , then x is equal to

Question 17. Solve $\tan ^{-1}(\frac{x}{y})-\tan ^{-1}(\frac{x-y}{x+y})$ is equal to

(A) π/2 (B) π/3 (C) π/4 (D) -3π/4