Question

Prove that $\frac{\pi }{2}-{\sin }^{-1}x=cose{c}^{-1}\frac{1}{\sqrt{1-x^2}}$

Answer 1

Let, ${\sin }^{-1}x=\theta$

or, $x=\sin \theta$

or, $\cos \theta =\sqrt{1-x^2}$

or, $\sin (\pi /2-\theta )=\sqrt{1-x^2}$

or, $\csc (\pi /2-\theta )=\left(\frac{1}{\sqrt{1-x^2}}\right)$

or, $\pi /2-\theta ={\csc }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)$

or, $\pi /2-{\sin }^{-1}x={\csc }^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right)$.