Question

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 -

• A. $[-1,1]-\{0\}$
• B. $(0,1]\cup \{-1\}$
• C. $[-1,0)\cup \{1\}$
• D. $[-1,1]$

Answer 1

Answer: (C)

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