In a triangle ABC,(b^2-c^2) cot A + (c^2-a^2) | vedclass

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(B) In $	riangle ABC$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these into the expression $(b^2-c^2) \cot A + (c^2-a^2

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$2R^2(\sin 2A - \sin 2B)$

Vedclass · Answer

(B) In $	riangle ABC$,we have $a = 2R \sin A$,$b = 2R \sin B$,and $c = 2R \sin C$.Substituting these into the expression $(b^2-c^2) \cot A + (c^2-a^2) \cot B$:$= 4R^2(\sin^2 B - \sin^2 C) \cot A + 4R^2(\sin^2 C - \sin^2 A) \cot B$$= 4R^2[\sin(B+C)\sin(B-C) \frac{\cos A}{\sin A} + \sin(C+A)\sin(C-A) \frac{\cos B}{\sin B}]$Since $A+B+C = \pi$,we have $\sin(B+C) = \sin A$ and $\sin(C+A) = \sin B$.$= 4R^2[\sin A \sin(B-C) \frac{\cos A}{\sin A} + \sin B \sin(C-A) \frac{\cos B}{\sin B}]$$= 4R^2[\sin(B-C)\cos A + \sin(C-A)\cos B]$$= 4R^2[\sin(B-C)(-\cos(B+C)) + \sin(C-A)(-\cos(C+A))]$$= 2R^2[-(2\sin(B-C)\cos(B+C)) - (2\sin(C-A)\cos(C+A))]$$= 2R^2[-(\sin 2B - \sin 2C) - (\sin 2C - \sin 2A)]$$= 2R^2[\sin 2A - \sin 2B]$

In a $\triangle ABC$,$(b^2-c^2) \cot A + (c^2-a^2) \cot B =$

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