Question

Prove $si{n}^{-1}\left(x\right)+co{s}^{-1}\left(x\right)=\left(\frac{\mathrm{\pi }}{2}\right)$

Answer 1

Prove that : $si{n}^{-1}\left(x\right)+co{s}^{-1}\left(x\right)=\left(\frac{\mathrm{\pi }}{2}\right)$

Let
$$\begin{gathered} \theta ={\sin }^{-1}x \\ \Rightarrow \sin \theta =x \\ \Rightarrow \sin \theta =\cos (\frac{\pi }{2}-\theta ) \\ \Rightarrow x=\cos (\frac{\pi }{2}-\theta ) \\ \Rightarrow \frac{\pi }{2}-\theta ={\cos }^{-1}x \\ \Rightarrow \frac{\pi }{2}-{\sin }^{-1}x={\cos }^{-1}x \\ \Rightarrow {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2} \end{gathered}$$

Hence, the given question is proved.