Problem
Evaluate: \( \cot\left(\sin^{-1}\left(\frac{3}{4}\right) + \sec^{-1}\left(\frac{4}{3}\right)\right) \)
Solution
Let:
\[ A = \sin^{-1}\left(\frac{3}{4}\right), \quad B = \sec^{-1}\left(\frac{4}{3}\right) \]
Step 1: For A
\[ \sin A = \frac{3}{4} = \frac{\text{Perpendicular}}{\text{Hypotenuse}} \]
- Perpendicular = 3
- Hypotenuse = 4
Base:
\[ \sqrt{4^2 – 3^2} = \sqrt{7} \]
\[ \tan A = \frac{3}{\sqrt{7}} \]
Step 2: For B
\[ \sec B = \frac{4}{3} \Rightarrow \cos B = \frac{3}{4} \]
So,
\[ B = \cos^{-1}\left(\frac{3}{4}\right) \]
Step 3: Key Insight
\[ A = \sin^{-1}\left(\frac{3}{4}\right), \quad B = \cos^{-1}\left(\frac{3}{4}\right) \]
\[ A + B = \frac{\pi}{2} \]
Step 4: Evaluate cot
\[ \cot\left(\frac{\pi}{2}\right) = 0 \]
Final Answer
\[ \boxed{0} \]
Explanation
sin⁻¹x + cos⁻¹x = π/2, hence cot(π/2) = 0.