Question
In triangle ABC, if \( C = 90^\circ \), find:
\[ \tan^{-1}\left(\frac{a}{b+c}\right) + \tan^{-1}\left(\frac{b}{c+a}\right) \]
Solution
In a right triangle at C:
\[ c^2 = a^2 + b^2 \]
Use identity:
\[ \tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \]
Let
\[ x = \frac{a}{b+c}, \quad y = \frac{b}{c+a} \]
Compute numerator:
\[ x + y = \frac{a}{b+c} + \frac{b}{c+a} = \frac{a(c+a) + b(b+c)}{(b+c)(c+a)} \]
\[ = \frac{ac + a^2 + b^2 + bc}{(b+c)(c+a)} \]
\[ = \frac{c(a+b) + (a^2 + b^2)}{(b+c)(c+a)} \]
Since \( a^2 + b^2 = c^2 \):
\[ = \frac{c(a+b+c)}{(b+c)(c+a)} \]
Compute denominator:
\[ 1 – xy = 1 – \frac{ab}{(b+c)(c+a)} \]
\[ = \frac{(b+c)(c+a) – ab}{(b+c)(c+a)} \]
\[ = \frac{bc + ca + c^2}{(b+c)(c+a)} \]
\[ = \frac{c(a+b+c)}{(b+c)(c+a)} \]
Thus:
\[ \frac{x+y}{1-xy} = 1 \]
\[ \tan^{-1}(1) = \frac{\pi}{4} \]
Final Answer:
\[ \boxed{\frac{\pi}{4}} \]
Key Concept
Use Pythagoras and tangent addition identity cleverly.