In a ` A B C ,` , if C is a right angle, then `tan^(-1)(a/(b+c))+tan^(-1)(b/(c+a))=` `pi/3` (b) `pi/4` (c) `(5pi)/2` (d) `pi/6`

Explanation

To solve the problem, we need to evaluate the expression:

\[ \tan^{-1}\left(\frac{a}{b+c}\right) + \tan^{-1}\left(\frac{b}{c+a}\right) \]

Given that \( C \) is a right angle in triangle \( ABC \), we can use the formula for the sum of two inverse tangents:

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