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Mathematics Question on Inverse Trigonometric Functions

If sin⁡−1 x+sin⁡−1 y=π2\sin^{-1} \, x + \sin^{-1} \,y= \frac{\pi}{2} , then x2x^2 is equal to

A

1−y21 - y^2

B

y2y^2

C

00

D

1−y\sqrt{1 - y }

Answer

1−y21 - y^2

Explanation

Solution

We have, sin⁡−1x+sin⁡−1y=π2\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}
⇒sin⁡−1x=π2−sin⁡−1y\Rightarrow \sin ^{-1} x=\frac{\pi}{2}-\sin ^{-1} y
⇒sin⁡−1x=cos⁡−1y\Rightarrow \sin ^{-1} x=\cos ^{-1} y
[∵sin⁡−1x+cos⁡−1x=π2]\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]
⇒sin⁡−1x=sin⁡−1(1−y2)\Rightarrow \sin ^{-1} x=\sin ^{-1}\left(\sqrt{1-y^{2}}\right)
⇒x=1−y2\Rightarrow x=\sqrt{1-y^{2}}
⇒x2=1−y2\Rightarrow x^{2}=1-y^{2}