Problem
Solve: \( \tan^{-1}\left(\frac{x-2}{x-4}\right) + \tan^{-1}\left(\frac{x+2}{x+4}\right) = \frac{\pi}{4} \)
Solution
Let:
\[ A = \tan^{-1}\left(\frac{x-2}{x-4}\right), \quad B = \tan^{-1}\left(\frac{x+2}{x+4}\right) \]
Step 1: Use tan(A + B)
\[ \tan(A + B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
Step 2: Substitute
\[ = \frac{\frac{x-2}{x-4} + \frac{x+2}{x+4}}{1 – \frac{(x-2)(x+2)}{(x-4)(x+4)}} \]
Step 3: Simplify numerator
\[ = \frac{(x-2)(x+4) + (x+2)(x-4)}{(x-4)(x+4)} \]
\[ = \frac{x^2 + 2x – 8 + x^2 – 2x – 8}{x^2 – 16} = \frac{2x^2 – 16}{x^2 – 16} \]
Step 4: Simplify denominator
\[ 1 – \frac{x^2 – 4}{x^2 – 16} = \frac{x^2 – 16 – (x^2 – 4)}{x^2 – 16} = \frac{-12}{x^2 – 16} \]
Step 5: Final tan value
\[ \tan(A + B) = \frac{2(x^2 – 8)}{-12} = -\frac{x^2 – 8}{6} \]
Step 6: Compare with RHS
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]
\[ -\frac{x^2 – 8}{6} = 1 \]
Step 7: Solve
\[ x^2 – 8 = -6 \]
\[ x^2 = 2 \]
\[ x = \pm \sqrt{2} \]
Step 8: Domain check
Expression undefined at \( x = \pm 4 \). Both \( \pm\sqrt{2} \) are valid.
Final Answer
\[ \boxed{x = \pm \sqrt{2}} \]
Explanation
Using tan(A+B) identity and simplifying rational expressions gives a quadratic.