Question

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

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

Answer 1

Answer: (D)

$\left\{1\right\}$

${\sin }^{-1}\sqrt{1-x}+{\cos }^{-1}x={\cot }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)-{\sin }^{-1}x$
$\Rightarrow {\sin }^{-1}\sqrt{1-x}+\frac{\pi }{2}={\cot }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)$
$\Rightarrow \frac{\pi }{2}-{\cot }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=-{\sin }^{-1}\sqrt{1-x}$
$\Rightarrow {\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)=-{\sin }^{-1}\sqrt{1-x}$
Thus is only true when, $x=\left\{1\right\}$