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With usual notations in $\Delta \mathrm{ABC}$, if $C=90^{\circ}$, then $\tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)=$
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The correct answer is:
$\frac{\pi}{4}$
$\therefore \quad \tan ^{-1}\left(\frac{a}{b+c}\right)+\tan ^{-1}\left(\frac{b}{c+a}\right)$
$=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right]$
$=\tan ^{-1}\left[\frac{a c+a^{2}+b c+b^{2}}{a c+b c+a b+c^{2}-a b}\right]=\tan ^{-1}\left(\frac{a c+b c+c^{2}}{a c+b c+c^{2}}\right) \quad \ldots\left[\because a^{2}+b^{2}=c^{2}\right.$, refer diagram $]$
$=\tan ^{-1}(1)=\frac{\pi}{4}$
$=\tan ^{-1}\left[\frac{\frac{a}{b+c}+\frac{b}{c+a}}{1-\left(\frac{a}{b+c}\right)\left(\frac{b}{c+a}\right)}\right]$
$=\tan ^{-1}\left[\frac{a c+a^{2}+b c+b^{2}}{a c+b c+a b+c^{2}-a b}\right]=\tan ^{-1}\left(\frac{a c+b c+c^{2}}{a c+b c+c^{2}}\right) \quad \ldots\left[\because a^{2}+b^{2}=c^{2}\right.$, refer diagram $]$
$=\tan ^{-1}(1)=\frac{\pi}{4}$
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