Inverse Trigonometric Functions Class 12 Maths Important Questions Chapter 2

Inverse Trigonometric Functions Exercise 2.1

Find the principal values of the following.

Question 1. \(\sin ^{-1}\left(\frac{-1}{2}\right)\)

Solution:

Let \(\sin ^{-1}\left(\frac{-1}{2}\right)=\theta \Rightarrow \sin \theta=-\frac{1}{2}\)

We know that the range of principal value of \(\sin ^{-1} \mathrm{x} is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\).

⇒ \(\sin \theta=-\frac{1}{2}=-\sin \frac{\pi}{6}=\sin \left(-\frac{\pi}{6}\right)  ( \sin (-\theta)=-\sin \theta)\)

⇒  \(\theta=-\frac{\pi}{6}\), where \(\theta \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \Rightarrow \sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)

Question 2. \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)\)

Solution:

Let \(\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\theta \Rightarrow \cos \theta=\frac{\sqrt{3}}{2}\)

We know that the range of principal value of \(\cos ^{-1} \mathrm{x} is [0, \pi]\).

∴  \(\cos \theta=\frac{\sqrt{3}}{2}=\cos \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6}\), where \(\theta \in[0, \pi] \Rightarrow \cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{6}\)

Question 3.\({cosec}^{-1}(2)\)

Solution:

Let \({cosec}^{-1}(2)=\theta \Rightarrow {cosec} \theta\)=2

We know that the range of principal value of \({cosec}^{-1} \mathrm{x} is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}\).

⇒  \({cosec} \theta=2={cosec} \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6}\), where \(\theta \in[-\frac{\pi}{2}\), \(\frac{\pi}{2}-{0}\)

⇒ \({cosec}^{-1}(2)\)

= \(\frac{\pi}{6}\)

Read and Learn More Class 12 Maths Chapter Wise with Solutions

Question 4. \(\tan ^{-1}(-\sqrt{3})\)

Solution:

Let \(\tan ^{-1}(-\sqrt{3})=\theta \Rightarrow \tan \theta=-\sqrt{3}\)

We know that the range of principal value of \(\tan ^{-1} \mathrm{x} is \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

⇒  \(\tan \theta=-\sqrt{3}=-\tan \frac{\pi}{3}=\tan \left(-\frac{\pi}{3}\right) (\tan (-\theta)=-\tan \theta)\)

⇒  \(\theta=-\frac{\pi}{3}\),where \(\theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \Rightarrow \tan ^{-1}(-\sqrt{3})=-\frac{\pi}{3}\)

Question 5. \(\cos ^{-1}\left(-\frac{1}{2}\right)\)

Solution:

Let \(\cos ^{-1}\left(-\frac{1}{2}\right)=\theta \Rightarrow \cos \theta=-\frac{1}{2}\)

We know that the range of principal value of \(\cos ^{-1} \mathrm{x} is [0, \pi]\)

⇒  \(\cos \theta=-\frac{1}{2}=-\cos \frac{\pi}{3}=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \frac{2 \pi}{3}(\cos (\pi-\theta)=-\cos \theta) \)

⇒  \(\theta=\frac{2 \pi}{3}\) ; where \(\theta \in[0, \pi] \Rightarrow \cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}\)

CBSE Class 12 Maths Chapter 2 Inverse Trigonometric Functions Important Question And Answers

Question 6. \(\tan ^{-1}(-1)\)

Solution:

Let \(\tan ^{-1}(-1)=\theta \Rightarrow \tan \theta\)=-1

We know that the range of principal value of \(\tan ^{-1} \mathrm{x} is \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

⇒  \(\tan \theta=-1=-\tan \frac{\pi}{4}=\tan \left(-\frac{\pi}{4}\right)\)

⇒  \((\tan (-\theta)=-\tan \theta)\)

⇒ \(\theta=-\frac{\pi}{4} ; where \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

∴ \(\tan ^{-1}(-1)=-\frac{\pi}{4}\)

Question 7. \(\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)\)

Solution:

Let \(\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\theta \Rightarrow \sec \theta=\frac{2}{\sqrt{3}}\)

We know that the range of principal value of \(\sec ^{-1} \mathrm{x} is [0, \pi]-\left\{\frac{\pi}{2}\right\}\)

∴ \(\sec \theta=\frac{2}{\sqrt{3}}=\sec \left(\frac{\pi}{6}\right) \Rightarrow \theta=\frac{\pi}{6} \text { where } \theta \in[0, \pi]-\left\{\frac{\pi}{2}\right\} \)

⇒  \(\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)=\frac{\pi}{6}\)

Question 8. \(\cot ^{-1}(\sqrt{3})\)

Solution:

Let \(\cot ^{-1}(\sqrt{3})=\theta \Rightarrow \cot \theta=\sqrt{3}\)

We know that the range of principal value of \(\cot ^{-1} \mathrm{x} is (0, \pi)\)

∴ \(\cot \theta=\sqrt{3}=\cot \frac{\pi}{6} \Rightarrow \theta=\frac{\pi}{6} ; where \theta \in(0, \pi) \Rightarrow \cot ^{-1}(\sqrt{3})=\frac{\pi}{6} \)

Question 9. \(\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)\)

Solution:

Let \(\cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\theta \Rightarrow \cos \theta=-\frac{1}{\sqrt{2}}\)

We know that the range of principal value of \(\cos ^{-1} \mathrm{x} is [0, \pi]\)

∴ \(\cos \theta=-\frac{1}{\sqrt{2}}=-\cos \frac{\pi}{4}=\cos \left(\pi-\frac{\pi}{4}\right)\) \([\cos (\pi-\theta)=-\cos \theta] \)

⇒  \(\theta=\frac{3 \pi}{4}\), where \(\theta \in[0, \pi] \Rightarrow \cos ^{-1}\left(-\frac{1}{\sqrt{2}}\right)=\frac{3 \pi}{4}\)

Question 10. \({cosec}^{-1}(-\sqrt{2})\)

Solution:

Let \({cosec}^{-1}(-\sqrt{2})=\theta \Rightarrow {cosec} \theta=-\sqrt{2}\)

We know that the range of principal value of \({cosec}^{-1} \mathrm{x} is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]-\{0\}\)

∴ \({cosec} \theta=-\sqrt{2}=-{cosec} \frac{\pi}{4}={cosec}\left(-\frac{\pi}{4}\right)\)      \(({cosec}(-\theta)=-{cosec} \theta) \)

⇒  \({cosec}^{-1}(-\sqrt{2})=-\frac{\pi}{4}\)

Question 11. \(\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)\)

Solution:

Let \(\tan ^{-1}(1)=x \Rightarrow \tan x=1=\tan \frac{\pi}{4} \Rightarrow x=\frac{\pi}{4}\)

where principal value of x \(\in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \tan ^{-1}(1)=\frac{\pi}{4}\)

Let \(\cos ^{-1}\left(-\frac{1}{2}\right)=y \Rightarrow \cos y=-\frac{1}{2}=-\cos \left(\frac{\pi}{3}\right)=\cos \left(\pi-\frac{\pi}{3}\right)=\cos \left(\frac{2 \pi}{3}\right)(\cos (\pi-\theta)=-\cos \theta)\)

⇒  y=\(\frac{2 \pi}{3}\), where principal value of y \(\in[0, \pi]\)

∴ \(\cos ^{-1}\left(-\frac{1}{2}\right)=\frac{2 \pi}{3}\)

Let \(\sin ^{-1}\left(-\frac{1}{2}\right)=z \Rightarrow \sin z=-\frac{1}{2}=-\sin \left(\frac{\pi}{6}\right)=\sin \left(-\frac{\pi}{6}\right)\)        (\(\sin (-\theta)=-\sin \theta)\)

⇒  z=-\(\frac{\pi}{6}\), where principal value of z \(\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

∴ \(\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}\)

∴ \(\tan ^{-1}(1)+\cos ^{-1}\left(-\frac{1}{2}\right)+\sin ^{-1}\left(-\frac{1}{2}\right)=x+y+z=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6}=\frac{3 \pi+8 \pi-2 \pi}{12}=\frac{9 \pi}{12}=\frac{3 \pi}{4}\)

Question 12. \(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)\)

Solution:

Let \(\cos ^{-1}\left(\frac{1}{2}\right)=x \Rightarrow \cos x=\frac{1}{2}=\cos \frac{\pi}{3}\)

⇒  \(\mathrm{x}=\frac{\pi}{3} \in[0, \pi]\) (Principal interval)

Again, let \(\sin ^{-1}\left(\frac{1}{2}\right)=y \Rightarrow \sin y=\frac{1}{2}=\sin \frac{\pi}{6}\)

⇒  \(\mathrm{y}=\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\) (Principal interval)

∴ \(\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)=x+2 y=\frac{\pi}{3}+\left(2 \times \frac{\pi}{6}\right)=\frac{2 \pi}{3}\)

Question 13. If \(\sin ^{-1}\) x=y, then

  1. 0 \(\leq \mathrm{y} \leq \pi\)
  2. –\(\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)
  3. \(0<y<\pi\)
  4. –\(\frac{\pi}{2}<y<\frac{\pi}{2}\)

Solution: 2. –\(\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\)

As range of \(\sin ^{-1} x is \left[-\frac{\pi}{2}, \frac{\pi}{2}\right], therefore -\frac{\pi}{2} \leq y \leq \frac{\pi}{2}\).

Question 14. \(\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)\) is equal to

  1. \(\pi\)
  2. –\(\frac{\pi}{3}\)
  3. \(\frac{\pi}{3}\)
  4. \(\frac{2 \pi}{3}\)

Solution: 2. –\(\frac{\pi}{3}\)

Let \(\tan ^{-1} \sqrt{3}=x \Rightarrow \tan x=\sqrt{3} \Rightarrow \tan x=\tan \frac{\pi}{3}\)

⇒ \(\mathrm{x}=\frac{\pi}{3} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\) (Principal interval)

Let \(\sec ^{-1}(-2)=y \Rightarrow \sec\) y=-2

⇒ \(\sec y=-\sec \frac{\pi}{3} \Rightarrow \sec y=\sec \left(\pi-\frac{\pi}{3}\right)\) [ \(\sec (\pi-\theta)=-\sec \theta\)]

⇒ \(\sec y=\sec \left(\frac{2 \pi}{3}\right) \Rightarrow y=\frac{2 \pi}{3} \in[0, \pi]-\left\{\frac{\pi}{2}\right\}\) (Principal interval)

∴ \(\tan ^{-1} \sqrt{3}-\sec ^{-1}(-2)=x-y=\frac{\pi}{3}-\frac{2 \pi}{3}=-\frac{\pi}{3}\)

Inverse Trigonometric Functions Exercise 2.2

Question 1. 3 \(\sin ^{-1} x=\sin ^{-1}\left(3 x-4 x^3\right), x \in\left[-\frac{1}{2}, \frac{1}{2}\right]\)

Solution:

Let \(\sin ^{-1} x=\theta \Rightarrow x=\sin \theta\), then

RHS =\(\sin ^{-1}\left(3 x-4 x^3\right)\)

= \(\sin ^{-1}\left[3 \sin \theta-4 \sin ^3 \theta\right]=\sin ^{-1}[\sin 3 \theta]=3 \theta \)      \([ \sin 3 \theta=3 \sin \theta-4 \sin ^3 \theta] \)

= 3 \(\sin ^{-1}\) x= LHS

Question 2. 3 \(\cos ^{-1} x=\cos ^{-1}\left(4 x^3-3 x\right), x \in\left[\frac{1}{2}, 1\right]\)

Solution:

Let \(\cos ^{-1} x=\theta \Rightarrow x=\cos \theta\), then

RHS =\(\cos ^{-1}\left(4 x^3-3 x\right)\)

= \(\cos ^{-1}\left[4 \cos ^3 \theta-3 \cos \theta\right]=\cos ^{-1}[\cos 3 \theta]=3 \theta\)      \([ \cos 3 \theta=4 \cos ^3 \theta-3 \cos \theta]\)

= 3 \(\cos ^{-1} x\)= LHS

Question 3. \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right), x \neq 0\)

Solution:

Put x=\(\tan \theta \Rightarrow \theta=\tan ^{-1}\)x

∴  \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{\sqrt{\sec ^2 \theta}-1}{\tan \theta}\right) \)        \(\left[ 1+\tan ^2 \theta=\sec ^2 \theta\right] \)

= \(\tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right]=\tan ^{-1}\left[\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right]=\tan ^{-1}\left[\frac{\frac{1-\cos \theta}{\cos \theta}}{\frac{\sin \theta}{\cos \theta}}\right]=\tan ^{-1}\left[\frac{1-\cos \theta}{\cos \theta} \times \frac{\cos \theta}{\sin \theta}\right]\)

= \(\tan ^{-1}[\frac{1-\cos \theta}{\sin \theta}]=\tan ^{-1}[\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}]\)

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

and \(\sin \theta=2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}\)

= \(\tan ^{-1}\left[\frac{2 \sin \frac{\theta}{2}}{2 \cos \frac{\theta}{2}}\right]=\tan ^{-1}\left[\tan \frac{\theta}{2}\right]=\frac{\theta}{2}=\frac{\tan ^{-1} x}{2}\) [from equation (1)]

∴ \(\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)=\frac{1}{2} \tan ^{-1} x\)

Question 4. \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right), 0<x<\pi \)

Solution:

⇒ \(\tan ^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)=\tan ^{-1}\left(\sqrt{\frac{2 \sin ^2(x / 2)}{2 \cos ^2(x / 2)}}\right)\) \((1-\cos x=2 \sin ^2(x / 2) and 1+\cos x=2 \cos ^2(x / 2)\)

= \(\tan ^{-1}\left(\sqrt{\tan ^2 \frac{x}{2}}\right)=\tan ^{-1}\left(\tan \frac{x}{2}\right)=\frac{x}{2}\)

Question 5. \(\tan ^{-1}\left(\frac{\cos x-\sin x}{\cos x+\sin x}\right), \frac{-\pi}{4}<x<\frac{3 \pi}{4} \)

Solution:

⇒ \(\tan ^{-1}(\frac{\cos x-\sin x}{\cos x+\sin x})=\tan ^{-1}(\frac{\frac{\cos x}{\cos x}-\frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x}+\frac{\sin x}{\cos x}})\) (inside the bracket divide numerator and denominator by \(\cos x)\)

= \(\tan ^{-1}\left(\frac{1-\tan x}{1+\tan x}\right)=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-x\right)\right]=\frac{\pi}{4}-x \left[\tan \left(\frac{\pi}{4}-x\right)=\frac{\tan \frac{\pi}{4}-\tan x}{1+\tan \frac{\pi}{4} \cdot \tan x}=\frac{1-\tan x}{1+\tan x}\right]\)

Question 6. \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right),|x|<a\)

Solution:

Put x=a \(\sin \theta \Rightarrow \frac{x}{a}=\sin \theta \Rightarrow \sin ^{-1}\left(\frac{x}{a}\right)=\theta\)

∴ \(\tan ^{-1}\left(\frac{x}{\sqrt{a^2-x^2}}\right) =\tan ^{-1}\left(\frac{a \sin \theta}{\sqrt{a^2-a^2 \sin ^2 \theta}}\right) \)

= \(\tan ^{-1}\left(\frac{a \sin \theta}{a \sqrt{1-\sin ^2 \theta}}\right)=\tan ^{-1}\left(\frac{\sin \theta}{\cos \theta}\right) \)

= \(\tan ^{-1}(\tan \theta)=\theta=\sin ^{-1}\left(\frac{x}{a}\right)\)

⇒ \(\sin ^2 x+\cos ^2 x=1\)

∴ \(\cos x=\sqrt{1-\sin ^2 x}\) [from equation (1)]

Question 7. \(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right), a>0 ; \frac{-a}{\sqrt{3}}<x<\frac{a}{\sqrt{3}}\)

Solution:

Put \(\mathrm{x}=\mathrm{a} \tan \theta\)

⇒ \(\frac{\mathrm{x}}{\mathrm{a}}=\tan \theta \Rightarrow \theta=\tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\)

∴ \(\tan ^{-1}\left(\frac{3 a^2 x-x^3}{a^3-3 a x^2}\right)=\tan ^{-1}\left[\frac{3 a^2(a \tan \theta)-(a \tan \theta)^3}{a^3-3 a(a \tan \theta)^2}\right]=\tan ^{-1}\left[\frac{a^3\left(3 \tan \theta-\tan ^3 \theta\right)}{a^3\left(1-3 \tan ^2 \theta\right)}\right]\)

=\(\tan ^{-1}\left[\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right]=\tan ^{-1}(\tan 3 \theta)\left(\tan 3 \theta=\frac{3 \tan \theta-\tan ^3 \theta}{1-3 \tan ^2 \theta}\right)\)

=3 \(\theta=3 \tan ^{-1}\left(\frac{\mathrm{x}}{\mathrm{a}}\right)\) [From equation (1)]

Question 8. \(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)

Solution:

⇒ \(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)

= \(\tan ^{-1}\left[2 \cos \left\{2 \sin ^{-1}\left(\sin \frac{\pi}{6}\right)\right\}\right] \quad\left(\sin \frac{\pi}{6}=\frac{1}{2}\right)\)

= \(\tan ^{-1}\left[2 \times \frac{1}{2}\right]=\tan ^{-1}(1) \)

= \(\tan ^{-1}\left[2 \cos \left(2 \times \frac{\pi}{6}\right)\right]=\tan ^{-1}\left[2 \cos \frac{\pi}{3}\right]\)    \(\left(\cos \frac{\pi}{3}=\frac{1}{2}\right)\)

= \(\tan ^{-1}\left(\tan \frac{\pi}{4}\right)=\frac{\pi}{4} \)   \(\left(\tan \frac{\pi}{4}=1\right) \)

Question 9. \(\tan \frac{1}{2}\left[\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right],|x|<1, y \cdot>0 \text { and } x y<1\)

Solution:

= \(\tan ^{-1}\left[2 \times \frac{1}{2}\right]=\tan ^{-1}(1) \) \(\tan \frac{1}{2}\left[\sin ^{-1}\left(\frac{2 x}{1+x^2}\right)+\cos ^{-1}\left(\frac{1-y^2}{1+y^2}\right)\right]\)

Put x=\(\tan \theta and y=\tan \phi\)

Now, \(\tan \frac{1}{2}\left[\sin ^{-1}\left(\frac{2 \tan \theta}{1+\tan ^2 \theta}\right)+\cos ^{-1}\left(\frac{1-\tan ^2 \phi}{1+\tan ^2 \phi}\right)\right] \Rightarrow \tan \frac{1}{2}\left[\sin ^{-1}(\sin 2 \theta)+\cos ^{-1}(\cos 2 \phi)\right] \)

∴ \(\tan \frac{1}{2}[2 \theta+2 \phi] \Rightarrow \tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\frac{x+y}{1-x y}\)

Question 10. \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)\)

Solution:

⇒ \(\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) =\sin ^{-1}\left[\sin \left(\pi-\frac{\pi}{3}\right)\right]=\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \)

= \(\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)      \((\sin ^{-1}(\sin \theta)=\theta, \theta \in[-\frac{\pi}{2}, \frac{\pi}{2}])\)

Question 11. \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)\)

Solution:

⇒ \(\tan ^{-1}\left(\tan \frac{3 \pi}{4}\right)=\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{4}\right)\right]=\tan ^{-1}\left(-\tan \frac{\pi}{4}\right) {[-\tan \theta=\tan (-\theta)]}\)

= \(\tan ^{-1}\left[\tan \left(-\frac{\pi}{4}\right)\right]=-\frac{\pi}{4} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) {\left[ \tan ^{-1}(\tan \theta)=\theta, \theta \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right]}\)

Question 12. \(\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)

Solution:

⇒ \(\tan \left(\sin ^{-1} \frac{3}{5}+\cot ^{-1} \frac{3}{2}\right)\)

Put \(\sin ^{-1} \frac{3}{5}=\theta \Rightarrow \sin \theta=\frac{3}{5} and \cot ^{-1} \frac{3}{2}=\phi \Rightarrow \cot \phi=\frac{3}{2} \)

Thus, \(\tan (\theta+\phi)\)  → Equation 1

Now, \(\tan \theta=\frac{3}{4}\) and \(\tan \phi=\frac{2}{3}(\sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^2}})\) Since, \(\tan (\theta+\phi)=\frac{\tan \theta+\tan \phi}{1-\tan \theta \tan \phi}=\left[\frac{\frac{3}{4}+\frac{2}{3}}{1-\left(\frac{3}{4} \times \frac{2}{3}\right)}\right]=\frac{17}{6} \Rightarrow(\theta+\phi)=\tan ^{-1} \frac{17}{6} \ldots \)  → Equation 2

Putting the value from eq.(2) in (1), \(\tan \left(\tan ^{-1} \cdot \frac{17}{6}\right)=\frac{17}{6}\)

Question 13. \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)\) is equal to?

  1. \(\frac{7 \pi}{6}\)
  2. \(\frac{5 \pi}{6}\)
  3. \(\frac{\pi}{3}\)
  4. \(\frac{\pi}{6}\)

Solution: 2. \(\frac{5 \pi}{6}\)

⇒ \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(2 \pi-\frac{5 \pi}{6}\right)\right]\)

∴ \(\cos ^{-1}\left(\cos \frac{7 \pi}{6}\right)=\cos ^{-1}\left[\cos \left(\frac{5 \pi}{6}\right)\right]=\frac{5 \pi}{6} \in[0, \pi] \left[ \cos ^{-1}(\cos \theta)=\theta, \theta \in[0, \pi]\right]\)

Question 14. \(\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]\) is equal to

  1. \(\frac{1}{2}\)
  2. \(\frac{1}{3}\)
  3. \(\frac{1}{4}\)
  4. 1

Solution: 4.

⇒ \(\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\frac{1}{2}\right)\right]=\sin \left[\frac{\pi}{3}-\sin ^{-1}\left(-\sin \frac{\pi}{6}\right)\right] \quad\left[ \sin \frac{\pi}{6}=\frac{1}{2}\right][ \sin (-x)=-\sin x]\)

=\(\sin \left[\frac{\pi}{3}-\sin ^{-1}\left\{\sin \left(-\frac{\pi}{6}\right)\right\}\right]=\sin \left[\frac{\pi}{3}-\left(-\frac{\pi}{6}\right)\right]=\sin \left[\frac{\pi}{3}+\frac{\pi}{6}\right]=\sin \frac{\pi}{2}\)=1

Question 15. \(\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3})\) is equal to

  1. \(\pi\)
  2. –\(\frac{\pi}{2}\)
  3. zero
  4. 2 \(\sqrt{3}\)

Solution: 2. –\(\frac{\pi}{2}\)

⇒ \(\tan ^{-1}(\sqrt{3})-\cot ^{-1}(-\sqrt{3}) \)

= \(\tan ^{-1}(\sqrt{3})-\left[\pi-\cot ^{-1}(\sqrt{3})\right]\) \([\cot ^{-1}(-x)=\pi-\cot ^{-1} x] \)

= \(\tan ^{-1}(\sqrt{3})-\pi+\cot ^{-1}(\sqrt{3})=\frac{\pi}{3}-\pi+\frac{\pi}{6}=-\frac{\pi}{2}\)

Inverse Trigonometric Functions Miscellaneous Exercise

Question 1. \(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)\)

Solution:

⇒ \(\cos ^{-1}(\cos \frac{13 \pi}{6})=\cos ^{-1}[\cos (2 \pi+\frac{\pi}{6})]\)

= \(\cos ^{-1}\left[\cos \left(\frac{\pi}{6}\right)\right]=\frac{\pi}{6} \in[0, \pi]    [ \cos ^{-1}(\cos \theta)=\theta, \theta \in[0, \pi] and \cos (2 \pi+\theta)=\cos \theta]\)

Question 2. \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)\)

Solution:

⇒ \(\tan ^{-1}(\tan \frac{7 \pi}{6})=\tan ^{-1}[\tan (\pi+\frac{\pi}{6}) [\tan ^{-1}(\tan \theta)=\theta, \theta \in(-\frac{\pi}{2}, \frac{\pi}{2})\) and \(\tan (\pi+\theta)=\tan \theta]]\)

∴ \(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)=\tan ^{-1}\left[\tan \frac{\pi}{6}\right]=\frac{\pi}{6} \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\)

Question 3. 2 \(\sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{24}{7}\right)\)

solution:

LHS =\(2 \sin ^{-1}(\frac{3}{5})=\sin ^{-1}[2 \times \frac{3}{5} \sqrt{1-(\frac{3}{5})^2}]\)   [\( 2 \sin ^{-1} x=\sin ^{-1}(2 x \sqrt{1-x^2})]\)

= \(\sin ^{-1}[2 \times \frac{3}{5} \times \frac{4}{5}]=\sin ^{-1}(\frac{24}{25})=\tan ^{-1}[\frac{\frac{24}{25}}{\sqrt{1-(\frac{24}{25})^2}}] (\sin ^{-1} x=\tan ^{-1} \frac{x}{\sqrt{1-x^2}})\)

= \(\tan ^{-1}\left[\frac{\frac{24}{25}}{\sqrt{1-\frac{576}{625}}}\right]=\tan ^{-1}\left[\frac{24}{25} \times \frac{25}{7}\right]=\tan ^{-1}\left[\frac{24}{7}\right]\)= RHS. Hence, proved.

Question 4. \(\sin ^{-1}\left(\frac{8}{17}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{77}{36}\right)\)

Solution:

L.H.S. =\(\sin ^{-1}\left(\frac{8}{17}\right)+\sin ^{-1}\left(\frac{3}{5}\right)\)

Let \(\sin ^{-1} \frac{8}{17}=x \Rightarrow \sin x=\frac{8}{17}\)

Now, \(\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\left(\frac{8}{17}\right)^2}=\frac{15}{17}\)

Thus, \(\tan x=\frac{\sin x}{\cos x}=\frac{8}{15}\)

Similarly Let \(\sin ^{-1} \frac{3}{5}=y \Rightarrow \sin y=\frac{3}{5}\)

Now, \(\cos y=\sqrt{1-\sin ^2 y}=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}\)

Thus, \(\tan y=\frac{\sin y}{\cos y}=\frac{3}{4}\)

Now, \(\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=\left[\frac{\frac{8}{15}+\frac{3}{4}}{1-\frac{8}{15} \times \frac{3}{4}}\right]=\frac{\frac{32+45}{60}}{\frac{60-24}{60}}=\frac{77}{36}\)

Hence, \(\tan (x+y)=\frac{77}{36} \Rightarrow x+y=\tan ^{-1}\left(\frac{77}{36}\right)\)

Putting the values of x and y; we get :

∴ \(\sin ^{-1}\left(\frac{8}{17}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{77}{36}\right)\)= RHS Hence, proved.

Question 5. \(\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right)\)

Solution:

LHS =\(\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)\)

Let \(\cos ^{-1} \frac{4}{5}=x \Rightarrow \cos x=\frac{4}{5}\)

Now, \(\sin x=\sqrt{1-\cos ^2 x}=\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}\)

Let y=\(\cos ^{-1} \frac{12}{13} \Rightarrow \cos y=\frac{12}{13}\)

Now, \(\sin y=\sqrt{1-\left(\frac{12}{13}\right)^2}=\frac{5}{13}\)

We know that \(\cos (x+y)=\cos x \cos y-\sin x \sin y\)

⇒ \(\cos (x+y)=\left(\frac{4}{5} \times \frac{12}{13}\right)-\left(\frac{3}{5} \times \frac{5}{13}\right)=\frac{48}{65}-\frac{15}{65}=\frac{33}{65}\)

x+y =\(\cos ^{-1}\left(\frac{33}{65}\right)\)

Putting the values of x and y

⇒ \(\cos ^{-1}\left(\frac{4}{5}\right)+\cos ^{-1}\left(\frac{12}{13}\right)=\cos ^{-1}\left(\frac{33}{65}\right)\)=RHS

∴ Hence proved.

Question 6. \(\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1}\left(\frac{56}{65}\right)\)

Solution:

LHS =\(\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right) \)

Let \(\cos ^{-1}\left(\frac{12}{13}\right)=x \Rightarrow \cos x=\frac{12}{13}. Now, \sin x=\sqrt{1-\cos ^2 x}=\sqrt{1-\left(\frac{12}{13}\right)^2}=\frac{5}{13}\)

Let y=\(\sin ^{-1} \frac{3}{5} \Rightarrow \sin y=\frac{3}{5}. Now, \cos y=\sqrt{1-\sin ^2 y}=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}\)

We know that \(\sin (x+y)=\sin x \cos y+\cos x \sin y=\left(\frac{5}{13} \times \frac{4}{5}\right)+\left(\frac{12}{13} \times \frac{3}{5}\right)=\frac{20}{65}+\frac{36}{65}\)

⇒ \(\sin (x+y)=\frac{56}{65} \Rightarrow x+y=\sin ^{-1}\left(\frac{56}{65}\right)\)

Putting values of x and y

∴ \(\cos ^{-1}\left(\frac{12}{13}\right)+\sin ^{-1}\left(\frac{3}{5}\right)=\sin ^{-1}\left(\frac{56}{65}\right)\)= RHS

Question 7. \(\tan ^{-1}\left(\frac{63}{16}\right)=\sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)\)

Solution:

LHS =\(\sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)\)

Let \(\sin ^{-1}\left(\frac{5}{13}\right)=x \Rightarrow \sin x=\frac{5}{13}\). Now, \(\cos x=\sqrt{1-\sin ^2 x}=\sqrt{1-\left(\frac{5}{13}\right)^2}=\frac{12}{13}\)

⇒ \(\tan x=\frac{\sin x}{\cos x} \Rightarrow \tan x=\frac{\left(\frac{5}{13}\right)}{\left(\frac{12}{13}\right)}=\frac{5}{12}\)

Let \(\cos ^{-1}\left(\frac{3}{5}\right)=y \Rightarrow \cos y=\frac{3}{5}\). Now, \(\sin y=\sqrt{1-\cos ^2 y}=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}\)

Let \(\cos ^{-1}\left(\frac{3}{5}\right)=y \Rightarrow \cos y=\frac{3}{5}\). Now, \(\sin y=\sqrt{1-\cos ^2 y}=\sqrt{1-\left(\frac{3}{5}\right)^2}=\frac{4}{5}\)

⇒ \(\tan y=\frac{\sin y}{\cos y} \Rightarrow \tan y=\frac{\left(\frac{4}{5}\right)}{\left(\frac{3}{5}\right)}=\frac{4}{3}\)

Now, \(\tan (x+y)=\frac{\tan x+\tan y}{1-\tan x \tan y}=\left[\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12} \times \frac{4}{3}}\right]=\left[\frac{\frac{15+48}{12 \times 3}}{\frac{12 \times 3-20}{12 \times 3}}\right]=\left(\frac{63}{16}\right)\)

Hence, \(\tan (x+y)=\frac{63}{16} \Rightarrow x+y=\tan ^{-1}\left(\frac{63}{16}\right)\)

Putting the values of x and y; we get \(\sin ^{-1}\left(\frac{5}{13}\right)+\cos ^{-1}\left(\frac{3}{5}\right)=\tan ^{-1}\left(\frac{63}{16}\right)\)= R.H.S.

Question 8. Prove that : \(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right), x \in[0,1]\)

Solution:

LHS =\(\tan ^{-1} \sqrt{x}=\frac{1}{2}\left(2 \tan ^{-1} \sqrt{x}\right)=\frac{1}{2} \cos ^{-1}\left(\frac{1-(\sqrt{x})^2}{1+(\sqrt{x})^2}\right) \quad\left[2 \tan ^{-1} x=\cos ^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right]\)

= \(\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\)= RHS

Question 9. \(\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)\)

Solution:

LHS =\(\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)\) →     Equation 1

Now, we can write

⇒ \(\sqrt{1+\sin x}=\sqrt{\sin ^2 \frac{x}{2}+\cos ^2 \frac{x}{2}+2 \sin \frac{x}{2} \cos \frac{x}{2}}=\sqrt{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)^2}\)

= \(\left|\sin \frac{x}{2}+\cos \frac{x}{2}\right|=\sin \frac{x}{2}+\cos \frac{x}{2}\)

Similarly, we can get \(\sqrt{1-\sin x}=\left|\cos \frac{x}{2}-\sin \frac{x}{2}\right|=\cos \frac{x}{2}-\sin \frac{x}{2}\)

On substituting the above two values in Equation (1), we get

LHS = \(\cot ^{-1}\left(\frac{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)+\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}{\left(\sin \frac{x}{2}+\cos \frac{x}{2}\right)-\left(\cos \frac{x}{2}-\sin \frac{x}{2}\right)}\right)=\cot ^{-1}\left(\frac{2 \cos \frac{x}{2}}{2 \sin \frac{x}{2}}\right)=\cot ^{-1}\left(\cot \frac{x}{2}\right)=\frac{x}{2}\)=RHS

Question 10. \(\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x,-\frac{1}{\sqrt{2}} \leq x \leq 1\)

Solution:

Put x=\(\cos \theta \Rightarrow \cos ^{-1} x=\theta\)

L.H.S.= \(\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\tan ^{-1}\left(\frac{\sqrt{1+\cos \theta}-\sqrt{1-\cos \theta}}{\sqrt{1+\cos \theta}+\sqrt{1-\cos \theta}}\right)\)

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

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

(inside the bracket divide numerator and denominator by \(\cos (\theta / 2)\) )

= \(\tan ^{-1}\left(\tan \left(\frac{\pi}{4}-\frac{\theta}{2}\right)\right) {\left[ \tan (A-B)=\frac{\tan A-\tan B}{1+\tan A \tan B}\right]}\)

= \(\frac{\pi}{4}-\frac{\theta}{2}=\frac{\pi}{4}-\frac{1}{2} \cos ^{-1} x\)= R.HS.

Question 11. \(2 \tan ^{-1}(\cos x)=\tan ^{-1}(2 {cosec} x)\)

Solution:

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

⇒ \(\tan \left[2 \tan ^{-1}(\cos x)\right]=2 {cosec} x [ Let \tan ^{-1}(\cos x)=\theta]\)

⇒ \(\tan 2 \theta=2 {cosec}\) x

⇒ \(\frac{2 \tan \theta}{1-\tan ^2 \theta}=2 {cosec}\) x

⇒ \(\frac{2 \tan ^{-1}\left(\tan ^{-1} \cos x\right)}{1-\tan ^2\left(\tan ^{-1} \cos x\right)}=2 {cosec} x\)

∴ \(\frac{2 \cos x}{1-\cos ^2 x}=\frac{2}{\sin x} \Rightarrow \cot x=1 \Rightarrow \cot x=\cot \frac{\pi}{4} \Rightarrow x=\frac{\pi}{4}\)

Question 12. \(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x,(x>0)\)

Solution:

Given; \(\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2} \tan ^{-1} x \Rightarrow 2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)=\tan ^{-1} x \Rightarrow \tan \left\{2 \tan ^{-1}\left(\frac{1-x}{1+x}\right)\right\}\)=x

⇒ \(\tan 2 \theta=x \Rightarrow \frac{2 \tan \theta}{1-\tan ^2 \theta}\)=x

⇒ \(\frac{2 \tan [\tan ^{-1}(\frac{1-x}{1+x})]}{1-\tan ^2[\tan ^{-1}(\frac{1-x}{1+x})]}=x [\frac{\frac{2(1-x)}{(1+x)}}{\frac{(1+x)^2-(1-x)^2}{(1+x)^2}}]=x [\frac{2(1-x)(1+x)}{(1+x)^2-(1-x)^2}]\)=x

⇒ \({\left[\frac{2\left(1-x^2\right)}{1+x^2+2 x-1-x^2+2 x}\right]=x \Rightarrow\left[\frac{2\left(1-x^2\right)}{4 x}\right]=x}\)

⇒ \(\left(\frac{\left(1-x^2\right)}{2 x}\right)=x \Rightarrow \frac{1-x^2}{2 x}=x \Rightarrow 1-x^2=2 x^2 \Rightarrow 1=3 x^2 \Rightarrow x^2=\frac{1}{3} \Rightarrow x= \pm \frac{1}{\sqrt{3}}\)

x= \(\frac{1}{\sqrt{3}}\) [x>0 given, so we do not take x=-\(\frac{1}{\sqrt{3}}]\)

Question 13. \(\sin \left(\tan ^{-1} x\right) ;|x|<1\) is equal to

  1. \(\frac{x}{\sqrt{1-x^2}}\)
  2. \(\frac{1}{\sqrt{1-x^2}}\)
  3. \(\frac{1}{\sqrt{1+x^2}}\)
  4. \(\frac{x}{\sqrt{1+x^2}}\)

Solution: 4. \(\frac{x}{\sqrt{1+x^2}}\)

∴ \(\sin \left(\tan ^{-1} \mathrm{x}\right)=\sin \left[\sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\right)\right]=\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}} \left[ \tan ^{-1} \mathrm{x}=\sin ^{-1}\left(\frac{\mathrm{x}}{\sqrt{1+\mathrm{x}^2}}\right)\right]\)

Question 14. If \(\sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2}\), then x is equal to

  1. 0, \(\frac{1}{2}\)
  2. 1, \(\frac{1}{2}\)
  3. 0
  4. \(\frac{1}{2}\)

Solution: 3. 0

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

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

= \(\cos ^{-1}(1-x) \left[ \sin ^{-1}(1-x)+\cos ^{-1}(1-x)=\frac{\pi}{2}\right]\)

⇒ \(\cos \left(-2 \sin ^{-1} \mathrm{x}\right)=1-\mathrm{x}\)

[Multiplying both sides by \(\cos x\) ]

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

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

⇒ \(1-2\left[\sin \left(\sin ^{-1} x\right)\right]^2=1-x {\left[\sin ^2 x=(\sin x)^2\right]} \)

⇒ \(1-2 x^2=1-x \Rightarrow 2 x^2-x\)=0

⇒ x(2 x-1)=0 \(\Rightarrow x\)=0 or 2 x-1=0 \(\Rightarrow\) x=0 or \(\frac{1}{2}\)

∴ But x=\(\frac{1}{2}\) does not satisfy the given equation, so x=0

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