Question

If $\sin^{-1} x + \sin^{-1} y = \pi/2$, then $x^2$ is equal to

• A. $1 - y^2$
• B. $1 + y^2$
• C. $\sqrt{1 - y^2}$
• D. $\sqrt{1 + y^2}$

Hint:

Whenever you see inverse trig functions summing to $\pi/2$, immediately think of complementary identities like $\sin^{-1}\theta + \cos^{-1}\theta = \pi/2$ or $\tan^{-1}\theta + \cot^{-1}\theta = \pi/2$. This converts a sum into an equality, which is much easier to solve.

Answer 1

The Correct Option is A

To solve the given problem, we need to evaluate the expression \(x^2\) when \(\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}\). Let's follow the steps:

1. Given the equation \(\sin^{-1} x + \sin^{-1} y = \frac{\pi}{2}\), we know that this implies \(\sin^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(y)\).
2. Using the identity \(\sin(\frac{\pi}{2} - \theta) = \cos(\theta)\), we can deduce that \(x = \cos(\sin^{-1} y)\).
3. The expression \(\cos(\sin^{-1} y)\) can be simplified using the Pythagorean identity: \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)}\).
4. Here, substitute \(\theta = \sin^{-1} y\). Thus, we have \(\cos(\sin^{-1} y) = \sqrt{1 - y^2}\).
5. This means \(x = \sqrt{1 - y^2}\).
6. Finally, square both sides to find \(x^2\):
   \(x^2 = (\sqrt{1 - y^2})^2 = 1 - y^2\).

Therefore, the correct answer is \(1 - y^2\), matching the given option.