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Mathematics Question on Properties of Inverse Trigonometric Functions

The value of sin⁡−1cos⁡(sin−1x)+cos⁡−1sin⁡(cos⁡−1x)\sin^{-1} \cos (sin^{-1}x)+\cos^{-1} \sin(\cos^{-1}x) is

A

0

B

π4\frac{\pi}{4}

C

π2\frac{\pi}{2}

D

π\pi

Answer

π2\frac{\pi}{2}

Explanation

Solution

sin−1[cos(sin−1x)]+cos−1[sin(cos−1x)]sin^{-1} \left[ cos\left(sin^{-1}x\right)\right] + cos^{-1}\left[sin\left(cos^{-1}x\right)\right] =sin−1[sin(π2−sin−1x)]+cos−1[cos(π2−cos−1x)]= sin^{-1}\left[sin\left(\frac{\pi}{2} -sin^{-1}x\right)\right] + cos^{-1} \left[cos\left(\frac{\pi}{2}-cos^{-1}x\right)\right] =π2−sin−1x+π2−cos−1x= \frac{\pi}{2} -sin^{-1} x + \frac{\pi}{2}-cos^{-1}x =π−(sin−1x+cos−1x)= \pi-\left(sin^{-1}x +cos^{-1}x\right) =π−π2= \pi- \frac{\pi}{2} =π2= \frac{\pi}{2}