If A+B+C=frac{3 pi}{2},then 4 sin A sin B sin | vedclass

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(A) Given $A+B+C=\frac{3 \pi}{2} \ldots(1)$Consider the expression $E = 4 \sin A \sin B \sin C+\cos 2 A+\cos 2 B+\cos 2 C$.Using the identity $\cos 2A

Vedclass · Accepted Answer

$-\sin (A+B+C)$

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(A) Given $A+B+C=\frac{3 \pi}{2} \ldots(1)$Consider the expression $E = 4 \sin A \sin B \sin C+\cos 2 A+\cos 2 B+\cos 2 C$.Using the identity $\cos 2A + \cos 2B = 2 \cos(A+B) \cos(A-B)$ and $\cos 2C = 1 - 2 \sin^2 C$:$\cos 2A + \cos 2B + \cos 2C = 2 \cos(A+B) \cos(A-B) + 1 - 2 \sin^2 C$.From $(1)$,$A+B = \frac{3 \pi}{2} - C$,so $\cos(A+B) = \cos(\frac{3 \pi}{2} - C) = -\sin C$.Substituting this:$= 2(-\sin C) \cos(A-B) + 1 - 2 \sin^2 C$$= 1 - 2 \sin C [\cos(A-B) + \sin C]$Since $\sin C = \sin(\frac{3 \pi}{2} - (A+B)) = -\cos(A+B)$:$= 1 - 2 \sin C [\cos(A-B) - \cos(A+B)]$Using $\cos(A-B) - \cos(A+B) = 2 \sin A \sin B$:$= 1 - 2 \sin C [2 \sin A \sin B] = 1 - 4 \sin A \sin B \sin C$.Thus,$4 \sin A \sin B \sin C + \cos 2A + \cos 2B + \cos 2C = 1$.Checking the options:$-\sin(A+B+C) = -\sin(\frac{3 \pi}{2}) = -(-1) = 1$.Therefore,the correct option is $A$.

If $A+B+C=\frac{3 \pi}{2}$,then $4 \sin A \sin B \sin C+\cos 2 A+\cos 2 B+\cos 2 C=$

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