^-1(23)+^-1(76)+^-1(34) is equal to:

Question

$\sin^{-1}\left(\sin\dfrac{2\pi}{3}ight)+\cos^{-1}\left(\cos\dfrac{7\pi}{6}ight)+	an^{-1}\left(	an\dfrac{3\pi}{4}ight)$ is equal to:

Accepted Answer

$$\sin^{-1}\left(\sin\frac{2\pi}{3}\right)=\pi-\frac{2\pi}{3}=\frac{\pi}{3},\qquad \cos^{-1}\left(\cos\frac{7\pi}{6}\right)=2\pi-\frac{7\pi}{6}=\frac{5\pi}{6},$$

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

Sum $=\dfrac{\pi}{3}+\dfrac{5\pi}{6}-\dfrac{\pi}{4}=\dfrac{4\pi+10\pi-3\pi}{12}=\dfrac{11\pi}{12}$.

Solutions · 1