Question

The value of ${\sin }^{-1}\frac{1}{2}+{\cos }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{\sqrt{3}}$ is

$\left(a\right)\frac{2\pi }{3}\left(b\right)\frac{\pi }{2}$
$\left(c\right)\frac{3\pi }{4}\left(d\right)\frac{5\pi }{6}$

Answer 1

$\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$
$\therefore {\sin }^{-1}\frac{1}{2}+{\cos }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{2}+\frac{\pi }{6}$
$=\frac{2\pi }{3}$

Alternate Solution:
${\sin }^{-1}\frac{1}{2}+{\cos }^{-1}\frac{1}{2}+{\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{\pi }{6}+\frac{\pi }{3}+\frac{\pi }{6}$
$=\frac{2\pi }{3}$