Question

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