In a Delta ABC,if {b^2} + {c^2} = 3{a^2},then | vedclass

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(C) We know that $\cot B + \cot C = \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C} = \frac{\sin C \cos B + \cos C \sin B}{\sin B \sin C} = \frac{\sin(B

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(C) We know that $\cot B + \cot C = \frac{\cos B}{\sin B} + \frac{\cos C}{\sin C} = \frac{\sin C \cos B + \cos C \sin B}{\sin B \sin C} = \frac{\sin(B+C)}{\sin B \sin C}$.Since $A+B+C = \pi$,$\sin(B+C) = \sin A$.So,$\cot B + \cot C = \frac{\sin A}{\sin B \sin C}$.Using the Sine Rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,we have $\sin A = \frac{a}{2R}, \sin B = \frac{b}{2R}, \sin C = \frac{c}{2R}$.Thus,$\cot B + \cot C = \frac{a/2R}{(b/2R)(c/2R)} = \frac{2Ra}{bc}$.Also,$\cot A = \frac{\cos A}{\sin A} = \frac{(b^2+c^2-a^2)/(2bc)}{a/(2R)} = \frac{R(b^2+c^2-a^2)}{abc}$.Given $b^2+c^2 = 3a^2$,then $b^2+c^2-a^2 = 2a^2$.So,$\cot A = \frac{R(2a^2)}{abc} = \frac{2Ra}{bc}$.Therefore,$\cot B + \cot C - \cot A = \frac{2Ra}{bc} - \frac{2Ra}{bc} = 0$.

In a $\Delta ABC$,if ${b^2} + {c^2} = 3{a^2}$,then $\cot B + \cot C - \cot A = $

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