$sin^{-1}\begin{Bmatrix}cos(sin^{-1}x)\end{Bmatrix}+cos^{-1}\begin{Bmatrix}sin(cos^{-1}x)\end{Bmatrix}$ is equal to
Answer & explanation
Correct answer: option 2
We have,
$sin^{-1}\begin{Bmatrix}cos(sin^{-1}x)\end{Bmatrix}+cos^{-1}\begin{Bmatrix}sin(cos^{-1}x)\end{Bmatrix}$
$= sin^{-1}\begin{Bmatrix}cos\left(\frac{\pi}{2}-cos^{-1}x\right)\end{Bmatrix} +cos^{-1}\begin{Bmatrix}sin(cos^{-1}x)\end{Bmatrix}$
$= sin^{-1}\begin{Bmatrix}sin (cos^{-1}x)\end{Bmatrix}+cos^{-1} \begin{Bmatrix}sin(cos^{-1}x)\end{Bmatrix}=\frac{\pi}{2}$