In a triangle ABC,if frac{a}{b^2-c^2} + frac{ | vedclass

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(D) Given,$\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$.Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$:$\frac{2R \sin A}{4R^2(\sin^

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$\frac{\pi}{3}$

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(D) Given,$\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$.Using the sine rule $a = 2R \sin A$,$b = 2R \sin B$,$c = 2R \sin C$:$\frac{2R \sin A}{4R^2(\sin^2 B - \sin^2 C)} + \frac{2R \sin C}{4R^2(\sin^2 B - \sin^2 A)} = 0$$\frac{\sin A}{\sin(B+C)\sin(B-C)} + \frac{\sin C}{\sin(B+A)\sin(B-A)} = 0$Since $A+B+C = \pi$,$\sin(B+C) = \sin A$ and $\sin(B+A) = \sin C$:$\frac{\sin A}{\sin A \sin(B-C)} + \frac{\sin C}{\sin C \sin(B-A)} = 0$$\frac{1}{\sin(B-C)} + \frac{1}{\sin(B-A)} = 0$$\sin(B-A) = -\sin(B-C) = \sin(C-B)$$B-A = C-B$ or $B-A = \pi - (C-B)$$2B = A+C$. Since $A+B+C = \pi$,$A+C = \pi - B$.$2B = \pi - B$ $\Rightarrow 3B = \pi$ $\Rightarrow B = \frac{\pi}{3}$.

In a $\triangle ABC$,if $\frac{a}{b^2-c^2} + \frac{c}{b^2-a^2} = 0$,then $B$ is equal to

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