Question

Considering only the principal values, if $\mathrm{ta}n\left({\cos }^{-1}x\right)=\sin \left[co{t}^{-1}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\right]$ then $x=$

• 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

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

To find the value of $x$:

Let $co{t}^{-1}\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)=\theta$

$$\begin{gathered} \Rightarrow \mathrm{co}t\theta =\frac{1}{2} \\ \Rightarrow \sin \theta =\frac{2}{\sqrt{5}} \end{gathered}$$

Putting them in the given equation, we get

$$\begin{gathered} \mathrm{ta}n\left({\cos }^{-1}x\right)=\sin \theta \\ \Rightarrow \mathrm{ta}n\left({\cos }^{-1}x\right)=\frac{2}{\sqrt{5}} \end{gathered}$$

Let ${\cos }^{-1}x=\phi$

$x=\cos \phi$

Now, $\tan \phi =\frac{2}{\sqrt{5}}$

$\Rightarrow \cos \phi =\frac{\sqrt{5}}{3}$

Thus $x=\cos \phi =\frac{\sqrt{5}}{3}$

Hence the correct option is D.