Question
Evaluate:
\[ \tan^{-1}\left(\frac{a}{b}\right) – \tan^{-1}\left(\frac{a-b}{a+b}\right) \]
Solution
Use identity:
\[ \tan^{-1}x – \tan^{-1}y = \tan^{-1}\left(\frac{x – y}{1 + xy}\right) \]
Let
\[ x = \frac{a}{b}, \quad y = \frac{a-b}{a+b} \]
Then,
\[ \frac{x – y}{1 + xy} = \frac{\frac{a}{b} – \frac{a-b}{a+b}}{1 + \frac{a}{b}\cdot \frac{a-b}{a+b}} \]
Simplify numerator:
\[ \frac{a(a+b) – b(a-b)}{b(a+b)} = \frac{a^2 + ab – ab + b^2}{b(a+b)} = \frac{a^2 + b^2}{b(a+b)} \]
Simplify denominator:
\[ \frac{b(a+b) + a(a-b)}{b(a+b)} = \frac{ab + b^2 + a^2 – ab}{b(a+b)} = \frac{a^2 + b^2}{b(a+b)} \]
Thus,
\[ \frac{x – y}{1 + xy} = 1 \]
Therefore,
\[ \tan^{-1}(1) = \frac{\pi}{4} \]
Final Answer:
\[ \boxed{\frac{\pi}{4}} \]
Key Concept
Use the formula for difference of inverse tangents and simplify carefully.