Question

If \[ \sin^{-1} x + \cos^{-1} y = 0, \] then \( x^2 + y^2 \) is equal to:

• A. \( \frac{1}{2} \)
• B. \( \sqrt{2} \)
• C. 1
• D. \( \frac{1}{\sqrt{2}} \)
• E. 2

Hint:

When solving equations involving inverse trigonometric functions, use the relationships between sine and cosine functions, such as \( \sin^{-1} x + \cos^{-1} y = 0 \), to express one variable in terms of the other. This will help simplify the expression.

Answer 1

The Correct Option is C

We are given that: \[ \sin^{-1} x + \cos^{-1} y = 0. \] This implies that: \[ \cos^{-1} y = -\sin^{-1} x. \] Now, recall that: \[ \cos^{-1} y = \sin^{-1} (\sqrt{1 - y^2}), \]
and since \( \cos^{-1} y \) and \( \sin^{-1} x \) are inverses of each other, we conclude that \( x = \sqrt{1 - y^2} \).
Step 1: Now, we compute \( x^2 + y^2 \): \[ x^2 = 1 - y^2 \quad \Rightarrow \quad x^2 + y^2 = (1 - y^2) + y^2 = 1. \]
Thus, the value of \( x^2 + y^2 \) is \( 1 \).
Therefore, the correct answer is option (C).