Question

If $\cos^{-1}(x - 2) = \sin^{-1}(y + 1)$, then the variables $x$ and $y$ satisfy the equation}

• A. $x^2 + y^2 - 4x + 2y + 4 = 0$
• B. $x^2 + y^2 - 4x + 2y + 5 = 0$
• C. $x^2 + y^2 - 4x + 2y + 6 = 0$
• D. $x^2 + y^2 - 2x + y + 4 = 0$
• E. $x^2 + y^2 + 4x + 2y + 4 = 0$

Hint:

For equations involving different inverse trigonometric functions, setting the whole expression equal to $\theta$ and using circular identities is the most consistent method.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Let the given angle be $\theta$. Then express $\cos \theta$ and $\sin \theta$ in terms of $x$ and $y$, and use the identity $\cos^2 \theta + \sin^2 \theta = 1$.

Step 2: Detailed Explanation:
1. Let $\cos^{-1}(x - 2) = \sin^{-1}(y + 1) = \theta$.
2. From the definitions of inverse functions:
\[ \cos \theta = x - 2 \]
\[ \sin \theta = y + 1 \]
3. Use the fundamental trigonometric identity:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
4. Substitute the expressions for $x$ and $y$:
\[ (x - 2)^2 + (y + 1)^2 = 1 \]
5. Expand the squares:
\[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = 1 \]
6. Simplify the equation by combining constant terms:
\[ x^2 + y^2 - 4x + 2y + 5 = 1 \]
\[ x^2 + y^2 - 4x + 2y + 4 = 0 \]

Step 3: Final Answer:
The equation is $x^2 + y^2 - 4x + 2y + 4 = 0$.