Question

Find the value of ${\sec }^{-1}(-2)-{\sin }^{-1}\left(\frac{1}{2}\right)$

Answer 1

Knowing that the range of $se{c}^{-1}x$ is $[0,\frac{\pi }{2})\cup (\frac{\pi }{2},\pi ]$ and that of $si{n}^{-1}x$ is $[\frac{\pi }{2},\frac{\pi }{2}]$,

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

$si{n}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{6}$

$\therefore$ $se{c}^{-1}(-2)-si{n}^{-1}\left(\frac{1}{2}\right)$ $=\frac{2\pi }{3}-\frac{\pi }{6}=\frac{\pi }{2}$