If in a $\triangle A B C, \angle C=\frac{\pi}{2}$, then $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=$

• Correct answer
  A

  $\frac{\pi}{4}$
• B

  $\frac{\pi}{2}$
• C

  $\frac{\pi}{3}$
• D

  None of these

Solution

Given, $\angle C=\frac{\pi}{2} \Rightarrow c^{2}=a^{2}+b^{2}$ \{by pythagoras $\}$ Now, $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)$ $=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right]$ $=\tan ^{-1}\left[\frac{a c+a^{2}+b^{2}+b c}{b c+b a+c^{2}+c a-a b}\right)$

$=\tan ^{-1}\left(\frac{a c+c^{2}+b c}{b c+c^{2}+c a}\right)$ $=\tan ^{-1}(1)=\frac{\pi}{4}$

Hence, the correct answer is option: 'A'.