Question

Write the value of $\cos \left(\frac{{\tan }^{-1}x+{\cot }^{-1}x}{3}\right)$, when $x=-\frac{1}{\sqrt{3}}$.

Answer 1

Given, $\cos \left(\frac{{\tan }^{-1}x+{\cot }^{-1}x}{3}\right)$

When $x=\frac{-1}{\sqrt{3}}$

$=\cos \left(\frac{{\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)+{\cot }^{-1}\left(\frac{-1}{\sqrt{3}}\right)}{3}\right)$

We know that

${\tan }^{-1}(-\theta )=-{\tan }^{-1}\theta$ and

${\cot }^{-1}(-\theta )=\pi -{\tan }^{-1}\theta$

$=\cos \left(\frac{-{\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)+\pi -{\cot }^{-1}\left(\frac{1}{\sqrt{3}}\right)}{3}\right)$

$=\cos \left(\frac{-\frac{\pi }{6}+\pi -\frac{\pi }{3}}{3}\right)$

$=\cos \left(\frac{-\frac{\pi }{6}+\frac{6\pi }{6}-\frac{2\pi }{6}}{3}\right)$

$=\cos \left(\frac{\frac{-\pi +6\pi -2\pi }{6}}{3}\right)$

$=\cos \left(\frac{\frac{3\pi }{6}}{3}\right)$

$=\cos \left(\frac{\frac{\pi }{2}}{3}\right)$

$=\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$