Problem
Solve: \( \tan^{-1}\left(\frac{x-2}{x-1}\right) + \tan^{-1}\left(\frac{x+2}{x+1}\right) = \frac{\pi}{4} \)
Solution
Let:
\[ A = \tan^{-1}\left(\frac{x-2}{x-1}\right), \quad B = \tan^{-1}\left(\frac{x+2}{x+1}\right) \]
Step 1: Use tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
Step 2: Substitute
\[ = \frac{\frac{x-2}{x-1} + \frac{x+2}{x+1}}{1 – \frac{(x-2)(x+2)}{(x-1)(x+1)}} \]
Step 3: Simplify numerator
\[ = \frac{(x-2)(x+1) + (x+2)(x-1)}{(x-1)(x+1)} \]
\[ = \frac{x^2 – x – 2 + x^2 + x – 2}{x^2 – 1} = \frac{2x^2 – 4}{x^2 – 1} \]
Step 4: Simplify denominator
\[ 1 – \frac{x^2 – 4}{x^2 – 1} = \frac{x^2 – 1 – (x^2 – 4)}{x^2 – 1} = \frac{3}{x^2 – 1} \]
Step 5: Final tan value
\[ \tan(A + B) = \frac{2(x^2 – 2)}{3} \]
Step 6: Compare with RHS
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
\[ \frac{2(x^2 – 2)}{3} = 1 \]
Step 7: Solve
\[ 2x^2 – 4 = 3 \]
\[ 2x^2 = 7 \Rightarrow x^2 = \frac{7}{2} \]
\[ x = \pm \sqrt{\frac{7}{2}} \]
Step 8: Domain check
Expression undefined at \( x = \pm 1 \). Both values are valid.
Final Answer
\[ \boxed{x = \pm \sqrt{\frac{7}{2}}} \]
Explanation
Using tan(A+B) identity and simplifying rational expressions leads to a quadratic equation.