In triangle ABC, if angle C = frac{pi}{2}, th | vedclass

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(A) Given $\angle C = 90^{\circ}$, we have $a^2 + b^2 = c^2$.Let $S = 	an^{-1}\left(\frac{a}{b+c}ight) + 	an^{-1}\left(\frac{b}{c+a}ight) + 	an

Vedclass · Accepted Answer

$	an^{-1}\left(\frac{r_3}{r}ight)$

Vedclass · Answer

(A) Given $\angle C = 90^{\circ}$, we have $a^2 + b^2 = c^2$.Let $S = 	an^{-1}\left(\frac{a}{b+c}ight) + 	an^{-1}\left(\frac{b}{c+a}ight) + 	an^{-1}\left(\frac{c}{a+b}ight)$.First, consider $	an^{-1}\left(\frac{a}{b+c}ight) + 	an^{-1}\left(\frac{b}{c+a}ight) = 	an^{-1}\left(\frac{\frac{a}{b+c} + \frac{b}{c+a}}{1 - \frac{ab}{(b+c)(c+a)}}ight)$.$= 	an^{-1}\left(\frac{ac + a^2 + bc + b^2}{(b+c)(c+a) - ab}ight) = 	an^{-1}\left(\frac{ac + bc + c^2}{bc + ab + c^2 + ac - ab}ight) = 	an^{-1}\left(\frac{c(a+b+c)}{c(a+b+c)}ight) = 	an^{-1}(1) = \frac{\pi}{4}$.Thus, $S = \frac{\pi}{4} + 	an^{-1}\left(\frac{c}{a+b}ight) = 	an^{-1}(1) + 	an^{-1}\left(\frac{c}{a+b}ight) = 	an^{-1}\left(\frac{1 + \frac{c}{a+b}}{1 - \frac{c}{a+b}}ight)$.$= 	an^{-1}\left(\frac{a+b+c}{a+b-c}ight) = 	an^{-1}\left(\frac{2s}{2(s-c)}ight) = 	an^{-1}\left(\frac{s}{s-c}ight)$.Since $r = \frac{\Delta}{s}$ and $r_3 = \frac{\Delta}{s-c}$, we have $\frac{r_3}{r} = \frac{s}{s-c}$.Therefore, $S = 	an^{-1}\left(\frac{r_3}{r}ight)$.

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

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