Problem
Solve: \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\left(\frac{8}{31}\right) \)
Solution
Let:
\[ A = \tan^{-1}(x+1), \quad B = \tan^{-1}(x-1) \]
Step 1: Use tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ = \frac{(x+1) + (x-1)}{1 – (x+1)(x-1)} = \frac{2x}{1 – (x^2 – 1)} = \frac{2x}{2 – x^2} \]
Step 2: Compare with RHS
\[ \frac{2x}{2 – x^2} = \frac{8}{31} \]
Step 3: Solve
\[ 31(2x) = 8(2 – x^2) \]
\[ 62x = 16 – 8x^2 \]
\[ 8x^2 + 62x – 16 = 0 \]
\[ 4x^2 + 31x – 8 = 0 \]
\[ (4x – 1)(x + 8) = 0 \]
Step 4: Solutions
\[ x = \frac{1}{4}, \quad x = -8 \]
Step 5: Principal value check
For inverse tangent addition, the sum must lie in \( (-\frac{\pi}{2}, \frac{\pi}{2}) \).
Check:
- For \( x = \frac{1}{4} \): valid
- For \( x = -8 \): invalid (sum falls outside principal range)
Final Answer
\[ \boxed{\frac{1}{4}} \]
Explanation
Using tan(A+B) identity and rejecting the extraneous solution based on principal value.