In triangle ABC,if frac{1}{a+b} + frac{1}{c+a | vedclass

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(C) Given the equation: $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$ Multiply both sides by $(a+b+c)$: $\frac{a+b+c}{a+b} + \frac{a+b+c}{c+a} = 3

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$\frac{\sqrt{3}}{2}$

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(C) Given the equation: $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$ Multiply both sides by $(a+b+c)$: $\frac{a+b+c}{a+b} + \frac{a+b+c}{c+a} = 3$ $1 + \frac{c}{a+b} + 1 + \frac{b}{c+a} = 3$ $\frac{c}{a+b} + \frac{b}{c+a} = 1$ $c(c+a) + b(a+b) = (a+b)(c+a)$ $c^2 + ac + ab + b^2 = ac + a^2 + bc + ab$ $b^2 + c^2 - a^2 = bc$ Using the Cosine Rule: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ $\cos A = \frac{bc}{2bc} = \frac{1}{2}$ Since $\cos A = \frac{1}{2}$,$A = 60^{\circ}$ Therefore,$\sin A = \sin 60^{\circ} = \frac{\sqrt{3}}{2}$

In $\triangle ABC$,if $\frac{1}{a+b} + \frac{1}{c+a} = \frac{3}{a+b+c}$,then $\sin A$ is equal to

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