Question

Show that:

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

Answer 1

Solution :-

Let $si{n}^{-1}x=\theta$

$\Rightarrow x=sin\theta$

$\Rightarrow x=cos(\frac{\pi }{2}-\theta )$

$\Rightarrow co{s}^{-1}x=\frac{\pi }{2}-\theta$

$\Rightarrow co{s}^{-1}x=\frac{\pi }{2}-si{n}^{-1}x$

$\Rightarrow si{n}^{-1}x+co{s}^{-1}x=\frac{\pi }{2}$