Question

Find the value of ${\tan }^{-1}\left(\tan \frac{5\pi }{6}\right)+{\cos }^{-1}\left(\cos \frac{13\pi }{6}\right)$

Answer 1

$\tan \left(\frac{5\pi }{6}\right)=\tan (\pi -\frac{\pi }{6})$

$=-\tan \frac{\pi }{6}$

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

$\cos \left(13\frac{\pi }{6}\right)=\cos (2\pi +\frac{\pi }{6})$

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

$\therefore {\tan }^{-1}\left(\frac{-1}{\sqrt{3}}\right)+{\cos }^{-1}\left(\frac{\sqrt{3}}{2}\right)$

$=-\frac{\pi }{6}+\frac{\pi }{6}=0$.