Question

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

• A. $\sqrt{3}$
• B. $\frac{-1}{\sqrt{3}}$
• C. $\frac{1}{\sqrt{3}}$
• D. $-\sqrt{3}$
• E. $1$

Answer 1

Answer: (C)

Given,
$ \tan ^{-1}(-x)+\cos ^{-1}\left(-\frac{1}{2}\right)=\frac{\pi}{2}$
$\Rightarrow \,-\tan ^{-1}(x)+\left(\pi-\frac{\pi}{3}\right)=\frac{\pi}{2}$
$\Rightarrow \, -\tan ^{-1} x=\frac{\pi}{2}-\frac{2 \pi}{3}=-\frac{\pi}{6}$
$ \Rightarrow \,\tan ^{-1} x=\frac{\pi}{6}$
$\therefore \, x=\tan \frac{\pi}{6}$
$=\frac{1}{\sqrt{3}}$