If A+B+C=270^{circ},then cos 2A + cos 2B + co | vedclass

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(C) Given that $A+B+C=270^{\circ}$.We need to evaluate $\cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C$.Using the sum-to-product formula $\cos 2

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$1$

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(C) Given that $A+B+C=270^{\circ}$.We need to evaluate $\cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C$.Using the sum-to-product formula $\cos 2B + \cos 2C = 2 \cos(B+C) \cos(B-C)$,we get:$= \cos 2A + 2 \cos(B+C) \cos(B-C) + 4 \sin A \sin B \sin C$.Since $B+C = 270^{\circ} - A$,then $\cos(B+C) = \cos(270^{\circ} - A) = -\sin A$.Substituting this:$= \cos 2A + 2(-\sin A) \cos(B-C) + 4 \sin A \sin B \sin C$$= (1 - 2 \sin^2 A) - 2 \sin A \cos(B-C) + 4 \sin A \sin B \sin C$$= 1 - 2 \sin A [\sin A + \cos(B-C)] + 4 \sin A \sin B \sin C$.Since $\sin A = \sin(270^{\circ} - (B+C)) = -\cos(B+C)$:$= 1 - 2 \sin A [-\cos(B+C) + \cos(B-C)] + 4 \sin A \sin B \sin C$.Using the identity $-\cos(B+C) + \cos(B-C) = 2 \sin B \sin C$:$= 1 - 2 \sin A (2 \sin B \sin C) + 4 \sin A \sin B \sin C$$= 1 - 4 \sin A \sin B \sin C + 4 \sin A \sin B \sin C$$= 1$.

If $A+B+C=270^{\circ}$,then $\cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C =$

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