Question

If ${\sin }^{-1}x+{\cot }^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{2},$ then the value of $x$ is

• A. $0$
• B. $\frac{2}{\sqrt{5}}$
• C. $\frac{1}{\sqrt{5}}$
• D. $\frac{\sqrt{3}}{2}$

Answer 1

The correct option is C $\frac{1}{\sqrt{5}}$

We know that
${\cot }^{-1}\frac{1}{2}={\cos }^{-1}\frac{1}{\sqrt{5}}$
We get
${\sin }^{-1}x+{\cos }^{-1}\frac{1}{\sqrt{5}}=\frac{\pi }{2}\Rightarrow x=\frac{1}{\sqrt{5}}(\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2},|x|\le 1)$