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

• Correct answer
  A

  $\tan ^{-1}\left(\frac{r_3}{r}\right)$
• B

  $\tan ^{-1}\left(\frac{r_1+r_2}{r_3}\right)$
• C

  $\tan ^{-1}\left(\frac{1}{r}\right)$
• D

  $\tan ^{-1}\left(\frac{r_1+r_2+r_3}{r}\right)$

Solution

Given, $\angle C=90^{\circ} \quad \therefore a^2+b^2=c^2$
$\begin{aligned} & \text { Now, } \tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)+\tan ^{-1}\left(\frac{c}{a+b}\right) \\ & =\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\frac{a-b}{(b+c)(c+a)}}\right]+\tan ^{-1}\left(\frac{c}{a+b}\right) \\ & =\tan ^{-1}\left[\frac{a c+a^2+b^2+b c}{b c+c^2+a b+a c-a b}\right]+\tan ^{-1}\left(\frac{c}{a+b}\right) \\ & =\tan ^{-1}\left[\frac{a c+c^2+b^2}{b c+c^2+a c}\right]+\tan ^{-1}\left(\frac{c}{a+b}\right) \\ & =\tan ^{-1}+\tan ^{-1}\left(\frac{c}{a+b}\right)=\tan ^{-1}\left(\frac{1+\frac{c}{a+b}}{1-\frac{c}{a+b}}\right) \\ & =\tan ^{-1}\left(\frac{a+b+c}{a+b-c}\right)=\tan ^{-1} \frac{2 s}{2(s-c)} \\ & =\tan ^{-1}\left(\frac{s}{s-c}\right)=\tan ^{-1}\left(\frac{r_3}{r}\right)\left[r_3=\frac{\Delta}{s-c}, r=\frac{\Delta}{s}\right] \\ & \end{aligned}$

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