Question

If \( \sin^{-1} x + \sin^{-1} y = \frac{\pi}{2} \), then \( x^2 \) is equal to

• A. \( 1 - y^2 \)
• B. \( y^2 \)
• C. \( (1 - y)^2 \)
• D. \( (1 - \sqrt{y})^2 \)

Hint:

When \( \sin^{-1} x + \sin^{-1} y = \frac{\pi}{2} \), use the identity \( x = \sqrt{1 - y^2} \) to find the relationship between \( x^2 \) and \( y^2 \).

Answer 1

The Correct Option is A

Step 1: Using the given equation.
We are given that: \[ \sin^{-1} x + \sin^{-1} y = \frac{\pi}{2} \] From the identity \( \sin^{-1} x + \sin^{-1} y = \frac{\pi}{2} \), we know that: \[ \sin^{-1} x = \frac{\pi}{2} - \sin^{-1} y \] Therefore, \( x = \cos(\sin^{-1} y) \).
Step 2: Using the trigonometric identity.
Using the identity \( \cos(\sin^{-1} y) = \sqrt{1 - y^2} \), we have: \[ x = \sqrt{1 - y^2} \] Squaring both sides, we get: \[ x^2 = 1 - y^2 \]
Step 3: Conclusion.
Thus, \( x^2 = 1 - y^2 \). The correct answer is (1) \( 1 - y^2 \).

Answer 2

We're given \( \sin^{-1}x + \sin^{-1}y = \frac{\pi}{2} \) and need \( x^2 \) in terms of \( y \). Taking the cosine of both sides gives a direct algebraic route; let's derive the relationship and check it against each option.

Let \( A = \sin^{-1}x \) and \( B = \sin^{-1}y \), so \( A + B = \frac{\pi}{2} \). Taking cosine of both sides: \[ \cos(A+B) = \cos\left(\frac{\pi}{2}\right) = 0 \] Expanding: \[ \cos A \cos B - \sin A \sin B = 0 \] Since \( \sin A = x \), \( \sin B = y \), \( \cos A = \sqrt{1-x^2} \), \( \cos B = \sqrt{1-y^2} \): \[ \sqrt{1-x^2}\sqrt{1-y^2} = xy \] Squaring: \[ (1-x^2)(1-y^2) = x^2y^2 \implies 1 - x^2 - y^2 + x^2y^2 = x^2y^2 \implies x^2 = 1 - y^2 \]

1. \( 1-y^2 \): This matches exactly what the cosine-addition derivation produces.
2. \( y^2 \): This would require \( x^2 = y^2 \), only a special case (e.g. \( x=y=\frac{1}{\sqrt{2}} \)), not the general relationship.
3. \( (1-y)^2 \): Expanding gives \( 1-2y+y^2 \), which does not match \( 1-y^2=(1-y)(1+y) \) unless \( y=0 \).
4. \( (1-\sqrt{y})^2 \): Introduces a square root of \( y \) that doesn't arise anywhere in the derivation.

The cosine-addition-formula derivation confirms \( x^2 = 1-y^2 \) directly and rules out the other forms.

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