Problem
Solve: \( \tan^{-1}(2+x) + \tan^{-1}(2-x) = \tan^{-1}\left(\frac{2}{3}\right) \), where \( x < -\sqrt{3} \) or \( x > \sqrt{3} \)
Solution
Let:
\[ A = \tan^{-1}(2+x), \quad B = \tan^{-1}(2-x) \]
Step 1: Use tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{(2+x)+(2-x)}{1 – (2+x)(2-x)} = \frac{4}{1 – (4 – x^2)} \]
\[ = \frac{4}{x^2 – 3} \]
Step 2: Compare with RHS
\[ \tan(A + B) = \tan\left(\tan^{-1}\left(\frac{2}{3}\right)\right) = \frac{2}{3} \]
\[ \frac{4}{x^2 – 3} = \frac{2}{3} \]
Step 3: Solve
\[ 12 = 2(x^2 – 3) \]
\[ 12 = 2x^2 – 6 \]
\[ 2x^2 = 18 \Rightarrow x^2 = 9 \]
\[ x = \pm 3 \]
Step 4: Apply domain
Given \( x < -\sqrt{3} \) or \( x > \sqrt{3} \):
- \( x = 3 \) ✔ valid
- \( x = -3 \) ✔ valid
Final Answer
\[ \boxed{x = \pm 3} \]
Explanation
Using tan(A+B) identity and checking domain conditions gives both solutions.