With usual notations in $\Delta \mathrm{ABC}$, if $C=90^{\circ}$, then $ an ^{-1}\left(\frac{a}{b+c} ight)+ an ^{-1}\left(\frac{b}{c+a} ight)=$

Question

With usual notations in $\Delta \mathrm{ABC}$, if $C=90^{\circ}$, then $	an ^{-1}\left(\frac{a}{b+c}ight)+	an ^{-1}\left(\frac{b}{c+a}ight)=$

Aicharya · Accepted Answer

$	herefore \quad 	an ^{-1}\left(\frac{a}{b+c}ight)+	an ^{-1}\left(\frac{b}{c+a}ight)$ $=	an ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}ight)\left(\frac{b}{c+a}ight)}ight]$ $=	an ^{-1}\left[\frac{a c+a^{2}+b c+b^{2}}{a c+b c+a b+c^{2}-a b}ight]=	an ^{-1}\left(\frac{a c+b c+c^{2}}{a c+b c+c^{2}}ight) \quad \ldots\left[\because a^{2}+b^{2}=c^{2}ight.$, refer diagram $]$ $=	an ^{-1}(1)=\frac{\pi}{4}$

With usual notations in $\Delta \mathrm{ABC}$, if $C=90^{\circ}$, then $\tan…

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