Question

For a real $x$, the angle ${\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)$ equals

• A. $\mathrm{\pi }$
• B. $\frac{\mathrm{\pi }}{2}$
• C. $0$
• D. None of these

Answer 1

The correct option is B

$\frac{\mathrm{\pi }}{2}$

Explanation for correct options:

Step 1: Applying theorem

Given,${\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)$

Let, ${\sin }^{-1}\left(x\right)=P$

$\Rightarrow$ $x=\sin \left(P\right)$

We know,

$\sin \left(A\right)=\cos (\frac{\mathrm{\pi }}{2}-A)$

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

Step 2: Substituting the value of x in the given expression

We know, ${\sin }^{-1}(\sin (A))=A$

$\Rightarrow$${\sin }^{-1}\left(\sin \right(P\left)\right)+{\cos }^{-1}\left(\cos \right(\frac{\mathrm{\pi }}{2}-P\left)\right)$$=P+\frac{\mathrm{\pi }}{2}-P$

$=\frac{\mathrm{\pi }}{2}$

$\therefore$The value of the expression ${\sin }^{-1}\left(x\right)+{\cos }^{-1}\left(x\right)$is $\frac{\mathbf{\pi }}{\mathbf{2}}$

Hence, Option (B) is correct.