Question

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

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

Answer 1

The correct option is C $\frac{1}{\sqrt{3}}$

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

$\Rightarrow {\tan }^{-1}x=\frac{\pi }{6}$
Therefore, $x=\tan \frac{\pi }{6}=\frac{1}{\sqrt{3}}$