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

Step 1: Given Condition
\[ \sin^{-1} x + \sin^{-1} y = \frac{\pi}{2} \]
Step 2: Use Identity
We know the identity: \[ \sin^{-1} t + \cos^{-1} t = \frac{\pi}{2} \] Comparing, \[ \sin^{-1} x = \frac{\pi}{2} - \sin^{-1} y = \cos^{-1} y \]
Step 3: Convert to Algebraic Form
\[ \sin^{-1} x = \cos^{-1} y \Rightarrow x = \sqrt{1 - y^2} \]
Step 4: Square Both Sides
\[ x^2 = ( \sqrt{1 - y^2} )^2 \] \[ x^2 = 1 - y^2 \]
Step 5: Final Answer
\[ \boxed{x^2 = 1 - y^2} \]