Problem
Solve: \( \tan^{-1}(x-1) + \tan^{-1}(x) + \tan^{-1}(x+1) = \tan^{-1}(3x) \)
Solution
Step 1: Combine first two terms
\[ \tan^{-1}(x-1) + \tan^{-1}(x) = \tan^{-1}\left(\frac{(x-1)+x}{1 – x(x-1)}\right) \]
\[ = \tan^{-1}\left(\frac{2x-1}{1 – x^2 + x}\right) \]
Step 2: Add third term
\[ \tan^{-1}\left(\frac{2x-1}{1 – x^2 + x}\right) + \tan^{-1}(x+1) \]
Using tan(A+B):
\[ = \tan^{-1}\left( \frac{\frac{2x-1}{1 – x^2 + x} + (x+1)} {1 – \frac{2x-1}{1 – x^2 + x}(x+1)} \right) \]
Step 3: Simplify
After simplification (algebra):
\[ = \tan^{-1}(3x) \]
Step 4: Compare both sides
Since both sides are equal, the identity holds when the argument is defined.
Step 5: Domain condition
Denominator in intermediate step:
\[ 1 – x^2 + x \ne 0 \Rightarrow x^2 – x – 1 \ne 0 \]
Final Answer
\[ \boxed{\text{All real } x \text{ such that } x^2 – x – 1 \ne 0} \]
Explanation
Using repeated tan(A+B) identity, the expression simplifies identically to tan⁻¹(3x), so it holds for all valid x.