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Question

Mathematics Question on Inverse Trigonometric Functions

If sin⁡−1x+sin⁡−1y=π2\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2} , then cos⁡−1x+cos⁡−1y\cos^{-1} x + \cos^{-1} y is equal to

A

π2\frac{\pi}{2}

B

π4\frac{\pi}{4}

C

π\pi

D

3π4\frac{3 \pi}{4}

Answer

π2\frac{\pi}{2}

Explanation

Solution

Given sin⁡−1x+sin⁡−1y=π2\sin^{-1}x + \sin^{-1}y = \frac{\pi}{2}
we know that sin⁡−1x+cos⁡−1x=π2\sin^{-1} x+\cos^{-1}x = \frac{\pi}{2}
⇒sin⁡−1x=π2−cos⁡−1x\Rightarrow \sin^{-1}x =\frac{\pi}{2} -\cos^{-1}x
∴\therefore Equation (1) becomes. π2−cos⁡−1x+π2−cos⁡−1y=π2 \frac{\pi}{2} -\cos^{-1} x + \frac{\pi}{2} -\cos^{-1} y =\frac{\pi}{2}
⇒cos⁡−1x+cos⁡−1y=π2\Rightarrow \cos^{-1} x + \cos^{-1} y = \frac{\pi}{2}