In Delta ABC,({b^2} - {c^2})cot A + ({c^2} - | vedclass

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(A) Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C =

Vedclass · Accepted Answer

$0$

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(A) Using the sine rule,$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$,so $\sin A = \frac{a}{2R}$,$\sin B = \frac{b}{2R}$,and $\sin C = \frac{c}{2R}$.Using the cosine rule,$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$.Thus,$\cot A = \frac{\cos A}{\sin A} = \frac{b^2 + c^2 - a^2}{2bc} \cdot \frac{2R}{a} = \frac{R(b^2 + c^2 - a^2)}{abc}$.Substituting this into the expression:$({b^2} - {c^2})\cot A = ({b^2} - {c^2}) \cdot \frac{R(b^2 + c^2 - a^2)}{abc} = \frac{R}{abc} (b^4 - c^4 - a^2b^2 + a^2c^2)$.Summing the three terms:$\sum ({b^2} - {c^2})\cot A = \frac{R}{abc} [(b^4 - c^4 - a^2b^2 + a^2c^2) + (c^4 - a^4 - b^2c^2 + b^2a^2) + (a^4 - b^4 - c^2a^2 + c^2b^2)]$.All terms inside the bracket cancel out to $0$.Therefore,the result is $0$.

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

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