Question

If \( \sin^{-1}\left(\frac{3}{5}\right) + \cos^{-1}\left(\frac{12}{13}\right) = \sin^{-1}\alpha \), then \( \alpha = \)

• A. \( \frac{56}{65} \)
• B. \( \frac{61}{65} \)
• C. \( \frac{63}{65} \)
• D. \( \frac{62}{65} \)

Hint:

When adding an inverse sine and an inverse cosine, use \(\sin(A+B)\) after finding the missing trigonometric values. Always check the quadrant of the sum to ensure the principal value matches.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given an equation involving inverse trigonometric functions and need to find \(\alpha\).

Step 2: Key Formula or Approach: Let \(u = \sin^{-1}(3/5)\) and \(v = \cos^{-1}(12/13)\). Then \(\sin(u+v) = \sin u \cos v + \cos u \sin v\). Use right‑triangle definitions to find \(\cos u\) and \(\sin v\).

Step 3: Detailed Explanation: Given \(\sin u = \frac{3}{5}\), with \(u \in (0,\frac{\pi}{2})\) (since the value is positive and less than 1). Then \(\cos u = \sqrt{1-(\frac{3}{5})^2} = \sqrt{1-\frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}\). Given \(\cos v = \frac{12}{13}\), with \(v \in (0,\frac{\pi}{2})\) (positive cosine). Then \(\sin v = \sqrt{1-(\frac{12}{13})^2} = \sqrt{1-\frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}\). Now compute \(\sin(u+v) = \sin u \cos v + \cos u \sin v = \frac{3}{5} \cdot \frac{12}{13} + \frac{4}{5} \cdot \frac{5}{13} = \frac{36}{65} + \frac{20}{65} = \frac{56}{65}\). Since \(u+v\) lies in \((0,\pi/2)\) (approximately \(36.87^\circ + 22.62^\circ = 59.49^\circ\)), \(\sin^{-1}(\sin(u+v)) = u+v\). Hence \(\alpha = \frac{56}{65}\).

Step 4: Final Answer:
\(\alpha = \frac{56}{65}\), which is option (A).