Problem
Solve: \( \tan^{-1}(x+2) + \tan^{-1}(x-2) = \tan^{-1}\left(\frac{8}{79}\right), \quad x > 0 \)
Solution
Let:
\[ A = \tan^{-1}(x+2), \quad B = \tan^{-1}(x-2) \]
Step 1: Use tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{(x+2)+(x-2)}{1 – (x+2)(x-2)} = \frac{2x}{1 – (x^2 – 4)} = \frac{2x}{5 – x^2} \]
Step 2: Compare with RHS
\[ \frac{2x}{5 – x^2} = \frac{8}{79} \]
Step 3: Solve
\[ 79(2x) = 8(5 – x^2) \]
\[ 158x = 40 – 8x^2 \]
\[ 8x^2 + 158x – 40 = 0 \]
\[ 4x^2 + 79x – 20 = 0 \]
\[ (4x – 1)(x + 20) = 0 \]
Step 4: Solutions
\[ x = \frac{1}{4}, \quad x = -20 \]
Step 5: Apply condition
Given \( x > 0 \), so:
\[ x = \frac{1}{4} \]
Final Answer
\[ \boxed{\frac{1}{4}} \]
Explanation
Using tan(A+B) identity and rejecting invalid solution based on domain condition.