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  4. The solution set of the equation sin -1 √1-x2+ cos -1 x= cot -1((√1-x2/x))- sin -1 x is -

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Q. The solution set of the equation $\sin ^{-1} \sqrt{1-x^2}+\cos ^{-1} x=\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)-\sin ^{-1} x$ is -

Inverse Trigonometric Functions

Solution:

$\sin ^{-1} \sqrt{1-x^2}+\frac{\pi}{2}=\cot ^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$
Let $\theta=\sin ^{-1} x,-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}, x \neq 0, \theta \neq 0$
$\text { so } \sin ^{-1} \cos \theta+\frac{\pi}{2}=\cot ^{-1} \cot \theta $
$\sin ^{-1} \sin \left(\frac{\pi}{2}-\theta\right)+\frac{\pi}{2}=\cot ^{-1} \cot \theta$
Case I : If $0< \theta \leq \frac{\pi}{2}$
then $0< \frac{\pi}{2}-\theta< \frac{\pi}{2}$
$\Rightarrow \frac{\pi}{2}-\theta+\frac{\pi}{2}=\theta \Rightarrow \theta=\frac{\pi}{2}$
$\sin \theta=1=x$
Case II : If $-\frac{\pi}{2} \leq \theta<0$
$\Rightarrow 0<-\theta \leq \frac{\pi}{2} \Rightarrow \frac{\pi}{2} \leq \frac{\pi}{2}-\theta<\pi$
then $\pi-\left(\frac{\pi}{2}-\theta\right)+\frac{\pi}{2}=\pi+\theta$
$\pi+\theta=\pi+\theta$
$-\frac{\pi}{0} \leq \theta<0 \Rightarrow-1 \leq \sin \theta<0 \Rightarrow-1 \leq x<0$