If angle A = 90^circ in the triangle ABC,then | vedclass

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(C) Given that $\angle A = 90^\circ$ in $	riangle ABC$,by the Pythagorean theorem,we have $a^2 = b^2 + c^2$,where $a$ is the hypotenuse.We need to ev

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$\pi/4$

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(C) Given that $\angle A = 90^\circ$ in $	riangle ABC$,by the Pythagorean theorem,we have $a^2 = b^2 + c^2$,where $a$ is the hypotenuse.We need to evaluate $	an^{-1}\left(\frac{c}{a+b}ight) + 	an^{-1}\left(\frac{b}{a+c}ight)$.Using the formula $	an^{-1}(x) + 	an^{-1}(y) = 	an^{-1}\left(\frac{x+y}{1-xy}ight)$,we get:$= 	an^{-1}\left[ \frac{\frac{c}{a+b} + \frac{b}{a+c}}{1 - \left(\frac{c}{a+b}ight)\left(\frac{b}{a+c}ight)} ight]$$= 	an^{-1}\left[ \frac{c(a+c) + b(a+b)}{(a+b)(a+c) - bc} ight]$$= 	an^{-1}\left[ \frac{ac + c^2 + ab + b^2}{a^2 + ac + ab + bc - bc} ight]$Since $b^2 + c^2 = a^2$,the numerator becomes $ac + ab + a^2$.$= 	an^{-1}\left[ \frac{a^2 + ab + ac}{a^2 + ab + ac} ight]$$= 	an^{-1}(1) = \frac{\pi}{4}$.

If $\angle A = 90^\circ$ in the triangle $ABC$,then $\tan^{-1}\left(\frac{c}{a+b}\right) + \tan^{-1}\left(\frac{b}{a+c}\right) = $

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