Question

The value of ${\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left(\sin \left(-\frac{\pi }{2}\right)\right)$ is

• A. $\frac{\pi }{6}$
• B. $\frac{\pi }{12}$
• C. $-\frac{\pi }{12}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (C)

Given, ${\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left[\sin \left(\frac{-\pi }{2}\right)\right]$
$={\tan }^{-1}\left(-\frac{1}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)+{\tan }^{-1}\left(-\sin \frac{\pi }{2}\right)[\because \sin (-\theta )=-\sin \theta ]$
$=\frac{-\pi }{6}+\frac{\pi }{3}-{\tan }^{-1}\left(\sin \frac{\pi }{2}\right)[\because {\tan }^{-1}(-x)=-{\tan }^{-1}x]$
$=\frac{\pi }{6}-{\tan }^{-1}(1)=\frac{\pi }{6}-\frac{\pi }{4}=-\frac{\pi }{12}$