Question

If ${\sin }^{-1}x+co{t}^{-1}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{2}$ then $x$ is:

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

Answer 1

The correct option is B

$\frac{1}{\sqrt{5}}$

Explanation for the correct option:

Simplification of expression:

Given that, ${\sin }^{-1}x+co{t}^{-1}\left(\frac{1}{2}\right)=\frac{\mathrm{\pi }}{2}$

$\begin{array}{rcl}& & \begin{array}{c}\therefore si{n}^{-1}\left(x\right)=\frac{\pi }{2}-{\cot }^{-1}\left(\frac{1}{2}\right)\\ ={\tan }^{-1}\left(\frac{1}{2}\right)\left[\because {\tan }^{-1}\left(x\right)+{\cot }^{-1}\left(x\right)=\frac{\pi }{2}\right]\end{array}\end{array}$

$\begin{array}{c}\therefore {\sin }^{-1}x={\sin }^{-1}\left[\frac{\frac{1}{2}}{\sqrt{1+\frac{1}{4}}}\right]\\ ={\sin }^{-1}\left[\frac{\frac{1}{2}}{\sqrt{\frac{4+1}{4}}}\right]\\ ={\sin }^{-1}\left[\frac{\frac{1}{2}}{\sqrt{\frac{5}{4}}}\right]\\ ={\sin }^{-1}\left[\frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}}\right]\\ ={\sin }^{-1}\left[\frac{1}{2}\times \frac{2}{\sqrt{5}}\right]\\ ={\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)\mathbf{\left[}\begin{array}{l}\mathbf{\because }{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{=}{\mathbf{sin}}^{\mathbf{-}\mathbf{1}}\mathbf{\theta }\\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{=}\mathbf{\theta }\end{array}\mathbf{\right]}\end{array}$

$\therefore x=\frac{1}{\sqrt{5}}$

Hence, option B is correct.