Question

If $\tan \left({\cos }^{-1}\left(x\right)\right)=\sin \left({\cot }^{-1}\left(\frac{1}{2}\right)\right)$ then $x$ is equal to

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

Answer 1

The correct option is D

$\frac{\sqrt{5}}{3}$

Step 1. Solve for $\sin \left({\cot }^{-1}\left(\frac{1}{2}\right)\right)$

Given, $\tan \left({\cos }^{-1}\left(x\right)\right)=\sin \left({\cot }^{-1}\left(\frac{1}{2}\right)\right)$

Let ${\cot }^{-1}\left(\frac{1}{2}\right)=\theta$

$\begin{array}{rcl}& \Rightarrow & \cot \left(\theta \right)=\frac{1}{2}\\ & \Rightarrow & \tan \left(\theta \right)=2\\ & \Rightarrow & \sin \left(\theta \right)=\frac{2}{\sqrt{5}};\left[\because \tan \left(x\right)=\frac{\mathrm{Perpendicular}}{\mathrm{Base}}\mathrm{and}\sin \left(x\right)=\frac{\mathrm{Perpendicular}}{\sqrt{{\left(\mathrm{Perpendicular}\right)}^2+{\left(\mathrm{Base}\right)}^2}}\right]\\ & \Rightarrow & \theta ={\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)\\ \therefore co{t}^{-1}\left(\frac{1}{2}\right)& =& {\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)\\ & \Rightarrow & \sin \left(co{t}^{-1}\left(\frac{1}{2}\right)\right)=\sin \left({\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)\right)\\ & \Rightarrow & \sin \left(co{t}^{-1}\left(\frac{1}{2}\right)\right)=\frac{2}{\sqrt{5}}\end{array}$

Step 2. Find the value of $x$

Now,

$\begin{array}{rcl}\tan \left({\cos }^{-1}\left(x\right)\right)& =& \sin \left({\cot }^{-1}\left(\frac{1}{2}\right)\right)\\ & \Rightarrow & \tan \left({\cos }^{-1}\left(x\right)\right)=\frac{2}{\sqrt{5}}\\ & \Rightarrow & {\cos }^{-1}\left(x\right)={\tan }^{-1}\left(\frac{2}{\sqrt{5}}\right)\end{array}$

Let, $\tan \left(\theta \right)=\frac{2}{\sqrt{5}}$

$$\begin{gathered} \Rightarrow \cos \left(\theta \right)=\frac{\sqrt{5}}{3}\left[\because \tan \left(x\right)=\frac{\mathrm{Perpendicular}}{\mathrm{Base}}\mathrm{and}\cos \left(x\right)=\frac{\mathrm{Base}}{\sqrt{{\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perpendicular}\right)}^2}}\right] \\ \Rightarrow \theta ={\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right) \end{gathered}$$

$\begin{array}{rcl}\therefore {\cos }^{-1}\left(x\right)& =& {\cos }^{-1}\left(\frac{\sqrt{5}}{3}\right)\\ & \Rightarrow & x=\frac{\sqrt{5}}{3}\end{array}$

Hence, $x$ is $\frac{\sqrt{5}}{3}$, so, the correct option is (D).