Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

Given, ${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)+{\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)+{\tan }^{-1}\tan \left(\frac{3\pi }{4}\right)$

Now simplifying each term we get, 

${\sin }^{-1}\sin \left(\frac{2\pi }{3}\right)=\pi -\frac{2\pi }{3}=\frac{\pi }{3}$

${\cos }^{-1}\left(\cos \frac{7\pi }{6}\right)=2\pi -\frac{7\pi }{6}=\frac{5\pi }{6}$

${\tan }^{-1}\tan \left(\frac{3\pi }{4}\right)=\frac{3\pi }{4}-\pi =\frac{-\pi }{4}$

${\sin }^{-1}\left(\sin \frac{2\pi }{3}\right)+{\cos }^{-1}\cos \frac{7\pi }{6}+{\tan }^{-1}\tan \frac{3\pi }{4}$

$=\frac{\mathrm{\pi }}{3}+\frac{5\mathrm{\pi }}{6}-\frac{\mathrm{\pi }}{4}$

$=\frac{11\pi }{12}$