Question

In the interval $x\in [0,1]$ the value of ${\cos }^{-1}\sqrt{1-x}+{\sin }^{-1}\sqrt{1-x}$ is

• A. $0$
• B. $\begin{array}{c}\frac{\pi }{2}\end{array}$
• C. $1$
• D. $\begin{array}{c}\pi \end{array}$

Answer 1

The correct option is B $\begin{array}{c}\frac{\pi }{2}\end{array}$

$\sqrt{1-x}$ is defined for $x\in [0,1]$
Here, the argument of ${\cos }^{-1}$ and ${\sin }^{-1}$ is same.
$\begin{array}{c}{\cos }^{-1}\sqrt{1-x}+{\sin }^{-1}\sqrt{1-x}=\frac{\pi }{2}\end{array}$