Question

Assertion ($A$) lf $0<x<\frac{\pi }{2}$ then ${\sin }^{-1}\left(cosx\right)+{\cos }^{-1}\left(sinx\right)=\pi -2x$
Reason (R) ${\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x\forall x\in [0,1]$

• A. Both A and R are true and R is the correct explanation of A
• B. Both A and R are true but R is not correct explanation of A
• C. A is true but Ris false
• D. A is false but R is true

Answer 1

The correct option is A Both A and R are true and R is the correct explanation of A

Given

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

${\sin }^{-1}\left(\sin (\frac{\pi }{2}-x)\right)+{\cos }^{-1}\left(\cos (\frac{\pi }{2}-x)\right)$

$(\sin \theta =\cos (\frac{\pi }{2}-\theta ))$

$\frac{\pi }{2}-x+\frac{\pi }{2}-x$

$\pi -2x$

We know

${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$

${\cos }^{-1}x=\frac{\pi }{2}-{\sin }^{-1}x$