Q100 / 170 · Manipal (MET) 2020 · Mathematics
If $\sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$, then value of $x$ will be
- A$0$
- Correct answerB$\frac{1}{\sqrt{5}}$
- C$\frac{2}{\sqrt{5}}$
- D$\frac{\sqrt{3}}{2}$
Solution
Given, $\sin ^{-1} x+\cot ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{2}$
$\Rightarrow \quad \sin ^{-1} x=\frac{\pi}{2}-\cot ^{-1}\left(\frac{1}{2}\right)$
$\begin{aligned} & \Rightarrow \quad \sin ^{-1} x=\tan ^{-1}\left(\frac{1}{2}\right) \\ & {\left[\because \tan ^{-1} \theta+\cot ^{-1} \theta=\frac{\pi}{2}\right] }\end{aligned}$
$\begin{array}{ll}\Rightarrow & x=\sin \left\{\tan ^{-1}\left(\frac{1}{2}\right)\right\} \\ \Rightarrow & x=\sin \left\{\sin ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right\} \\ \Rightarrow & x=\frac{1}{\sqrt{5}}\end{array}$
Hence, the correct answer is option: 'B'.