With usual notations in Delta ABC,if C=90^{ci | vedclass

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Question

(A) Given that in $\Delta ABC$,$C=90^{\circ}$,therefore by Pythagoras theorem,$a^{2}+b^{2}=c^{2}$.We need to evaluate $	an ^{-1}\left(\frac{a}{b+c}

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(A) Given that in $\Delta ABC$,$C=90^{\circ}$,therefore by Pythagoras theorem,$a^{2}+b^{2}=c^{2}$.We need to evaluate $	an ^{-1}\left(\frac{a}{b+c}ight)+	an ^{-1}\left(\frac{b}{c+a}ight)$.Using the formula $	an ^{-1} x + 	an ^{-1} y = 	an ^{-1} \left( \frac{x+y}{1-xy} 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]$$= 	an ^{-1} \left[ \frac{a(c+a) + b(b+c)}{(b+c)(c+a) - ab} ight]$$= 	an ^{-1} \left[ \frac{ac + a^{2} + b^{2} + bc}{bc + ab + c^{2} + ac - ab} ight]$Since $a^{2}+b^{2}=c^{2}$,the numerator becomes $ac + c^{2} + bc = c(a+b+c)$.The denominator becomes $bc + ac + c^{2} = c(b+a+c)$.$= 	an ^{-1} \left( \frac{c(a+b+c)}{c(a+b+c)} ight)$$= 	an ^{-1}(1) = \frac{\pi}{4}$.

With usual notations in $\Delta ABC$,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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