Problem
Solve: \( \tan^{-1}\left(\frac{x}{2}\right) + \tan^{-1}\left(\frac{x}{3}\right) = \frac{\pi}{4}, \quad 0 < x < \sqrt{6} \)
Solution
Let:
\[ A = \tan^{-1}\left(\frac{x}{2}\right), \quad B = \tan^{-1}\left(\frac{x}{3}\right) \]
Step 1: Use tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{\frac{x}{2} + \frac{x}{3}}{1 – \frac{x}{2}\cdot\frac{x}{3}} = \frac{\frac{5x}{6}}{1 – \frac{x^2}{6}} \]
\[ = \frac{5x}{6 – x^2} \]
Step 2: Compare with RHS
\[ \tan(A + B) = \tan\left(\frac{\pi}{4}\right) = 1 \]
\[ \frac{5x}{6 – x^2} = 1 \]
Step 3: Solve
\[ 5x = 6 – x^2 \]
\[ x^2 + 5x – 6 = 0 \]
\[ (x+6)(x-1) = 0 \]
Step 4: Solutions
\[ x = -6, \quad x = 1 \]
Step 5: Apply domain
Given \( 0 < x < \sqrt{6} \), so:
\[ x = 1 \]
Final Answer
\[ \boxed{1} \]
Explanation
Using tan(A+B) identity and rejecting invalid solution based on given interval.