Question

If \[ \sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}=\frac{\pi}{2}, \] then \(x=\)

• A. \(12\)
• B. \(7\)
• C. \(13\)
• D. \(15\)

Hint:

When two inverse sine angles add to \(\frac{\pi}{2}\), use the complementary angle relation and the identity \(\sin^2A+\cos^2A=1\).

Answer 1

The Correct Option is C

Concept: If \[ \sin^{-1}a+\sin^{-1}b=\frac{\pi}{2}, \] then the corresponding angles are complementary. Hence: \[ a^2+b^2=1 \]

Step 1: Given: \[ \sin^{-1}\frac{5}{x}+\sin^{-1}\frac{12}{x}=\frac{\pi}{2} \]

Step 2: Let \[ \sin^{-1}\frac{5}{x}=A \] and \[ \sin^{-1}\frac{12}{x}=B \] Then: \[ A+B=\frac{\pi}{2} \] So, \[ B=\frac{\pi}{2}-A \]

Step 3: Therefore, \[ \sin B=\sin\left(\frac{\pi}{2}-A\right)=\cos A \]

Step 4: Since \[ \sin B=\frac{12}{x} \] and \[ \sin A=\frac{5}{x} \] We get: \[ \cos A=\frac{12}{x} \]

Step 5: Use the identity: \[ \sin^2A+\cos^2A=1 \] \[ \left(\frac{5}{x}\right)^2+\left(\frac{12}{x}\right)^2=1 \] \[ \frac{25}{x^2}+\frac{144}{x^2}=1 \] \[ \frac{169}{x^2}=1 \] \[ x^2=169 \] \[ x=13 \] Therefore, \[ \boxed{13} \]