Question

The solution set of the equation ${\sin }^{-1}\sqrt{1-x^2}+{\cos }^{-1}x={\cot }^{-1}\frac{\sqrt{1-x^2}}{x}-{\sin }^{-1}x$ is?

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

Answer 1

Answer: (A)

$[-1,1]-\left\{0\right\}$

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

let $x=\cos \theta ,\theta ={\cos }^{-1}x$

$$\begin{array}{c}\frac{\pi }{2}={\cot }^{-1}\frac{\sqrt{1-{\cos }^2\theta }}{\cos \theta }-{\sin }^{-1}\sqrt{1-{\cos }^2\theta }\\ \frac{\pi }{2}={\cot }^{-1}\left(\frac{\sin \theta }{\cos \theta }\right)-{\sin }^{-1}\left(\sin \theta \right)\\ \frac{\pi }{2}={\cot }^{-1}\left[\cot \left(\frac{\pi }{2}-\theta \right)\right]-\theta \\ \frac{\pi }{2}=\frac{\pi }{2}-\theta -\theta \\ 0=-2\theta \\ \theta =0\\ {\cos }^{-1}x=0\\ x=\cos 0^0\\ x=1\\ \left[-1,1\right]-\left\{0\right\}\end{array}$$