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Question: $(\sin^{-1}x)^2+(\sin^{-1}y)^2+2(\sin^{-1}x)(\sin^{-1}y)=\pi^2$, then $x^2+y^2$ is equal to:...

(sin⁡−1x)2+(sin⁡−1y)2+2(sin⁡−1x)(sin⁡−1y)=π2(\sin^{-1}x)^2+(\sin^{-1}y)^2+2(\sin^{-1}x)(\sin^{-1}y)=\pi^2, then x2+y2x^2+y^2 is equal to:

A

0

B

2

C

4

D

None of these

Answer

2

Explanation

Solution

The given equation is (sin⁡−1x)2+(sin⁡−1y)2+2(sin⁡−1x)(sin⁡−1y)=π2(\sin^{-1}x)^2+(\sin^{-1}y)^2+2(\sin^{-1}x)(\sin^{-1}y)=\pi^2. This can be rewritten as:

(sin⁡−1x+sin⁡−1y)2=π2(\sin^{-1}x + \sin^{-1}y)^2 = \pi^2

Taking the square root of both sides:

sin⁡−1x+sin⁡−1y=±π\sin^{-1}x + \sin^{-1}y = \pm \pi

Since the range of sin⁡−1u\sin^{-1}u is [−π2,π2][-\frac{\pi}{2}, \frac{\pi}{2}], we have −π2≤sin⁡−1x≤π2-\frac{\pi}{2} \le \sin^{-1}x \le \frac{\pi}{2} and −π2≤sin⁡−1y≤π2-\frac{\pi}{2} \le \sin^{-1}y \le \frac{\pi}{2}. Thus, −π≤sin⁡−1x+sin⁡−1y≤π-\pi \le \sin^{-1}x + \sin^{-1}y \le \pi.

Case 1: sin⁡−1x+sin⁡−1y=π\sin^{-1}x + \sin^{-1}y = \pi

This is only possible if sin⁡−1x=π2\sin^{-1}x = \frac{\pi}{2} and sin⁡−1y=π2\sin^{-1}y = \frac{\pi}{2}. Therefore, x=sin⁡(π2)=1x = \sin(\frac{\pi}{2}) = 1 and y=sin⁡(π2)=1y = \sin(\frac{\pi}{2}) = 1. In this case, x2+y2=12+12=2x^2 + y^2 = 1^2 + 1^2 = 2.

Case 2: sin⁡−1x+sin⁡−1y=−π\sin^{-1}x + \sin^{-1}y = -\pi

This is only possible if sin⁡−1x=−π2\sin^{-1}x = -\frac{\pi}{2} and sin⁡−1y=−π2\sin^{-1}y = -\frac{\pi}{2}. Therefore, x=sin⁡(−π2)=−1x = \sin(-\frac{\pi}{2}) = -1 and y=sin⁡(−π2)=−1y = \sin(-\frac{\pi}{2}) = -1. In this case, x2+y2=(−1)2+(−1)2=2x^2 + y^2 = (-1)^2 + (-1)^2 = 2.

In both cases, x2+y2=2x^2 + y^2 = 2.