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:
We have an equation relating an inverse cosine to an inverse sine.
If \(\cos^{-1} A = \sin^{-1} B = \theta\), then \(\cos \theta = A\) and \(\sin \theta = B\).
Step 2: Key Formula or Approach:
Use the fundamental trigonometric identity: \(\cos^2 \theta + \sin^2 \theta = 1\).
Therefore, \(A^2 + B^2 = 1\).
Step 3: Detailed Explanation:
Let \(\theta = \cos^{-1}(x - 2) = \sin^{-1}(y + 1)\).
This means \(\cos \theta = x - 2\) and \(\sin \theta = y + 1\).
Substitute these into \(\cos^2 \theta + \sin^2 \theta = 1\):
\[ (x - 2)^2 + (y + 1)^2 = 1 \] Expand the squared terms:
\[ (x^2 - 4x + 4) + (y^2 + 2y + 1) = 1 \] Combine all terms:
\[ x^2 + y^2 - 4x + 2y + 5 = 1 \] Subtract 1 from both sides:
\[ x^2 + y^2 - 4x + 2y + 4 = 0 \] Step 4: Final Answer:
The equation is \(x^2 + y^2 - 4x + 2y + 4 = 0\).