Problem
Find the value of \( \tan^{-1}\left(\frac{x}{y}\right) – \tan^{-1}\left(\frac{x-y}{x+y}\right) \)
Solution
Let:
\[ A = \tan^{-1}\left(\frac{x}{y}\right), \quad B = \tan^{-1}\left(\frac{x-y}{x+y}\right) \]
Step 1: Use tan(A − B)
\[ \tan(A – B) = \frac{\tan A – \tan B}{1 + \tan A \tan B} \]
Step 2: Substitute
\[ = \frac{\frac{x}{y} – \frac{x-y}{x+y}}{1 + \frac{x}{y}\cdot\frac{x-y}{x+y}} \]
Step 3: Simplify numerator
\[ = \frac{\frac{x(x+y) – y(x-y)}{y(x+y)}}{1 + \frac{x(x-y)}{y(x+y)}} \]
\[ = \frac{\frac{x^2 + xy – xy + y^2}{y(x+y)}}{\frac{y(x+y) + x(x-y)}{y(x+y)}} \]
\[ = \frac{x^2 + y^2}{x^2 + y^2} = 1 \]
Step 4: Final angle
\[ \tan(A – B) = 1 \Rightarrow A – B = \frac{\pi}{4} \]
Final Answer
\[ \boxed{\frac{\pi}{4}} \]
Explanation
Using tan(A−B) identity, the expression simplifies to 1, giving angle π/4.