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If $A+B+C=270^{\circ}$, then $\cos 2 A+\cos 2 B+\cos 2 C$ is equal to :
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Verified Answer
The correct answer is:
$1-4 \sin A \sin B \sin C$
$\cos 2 A+\cos 2 B+\cos 2 C$
$=2 \cos (A+B) \cos (A-B)+1-2 \sin ^2 C$
$=2 \cos \left(\frac{3 \pi}{2}-C\right) \cos (A-B)+1-2 \sin ^2 C$
$\left[\because A+B+C=270^{\circ} \Rightarrow B+C=\frac{3 \pi}{2}-C\right\rfloor$
$=1-2 \sin C[\cos (A-B)+\sin C]$
$=1-2 \sin C[\cos (A-B)-\cos (A+B)]$
$=1-4 \sin A \sin B \sin C$
$=2 \cos (A+B) \cos (A-B)+1-2 \sin ^2 C$
$=2 \cos \left(\frac{3 \pi}{2}-C\right) \cos (A-B)+1-2 \sin ^2 C$
$\left[\because A+B+C=270^{\circ} \Rightarrow B+C=\frac{3 \pi}{2}-C\right\rfloor$
$=1-2 \sin C[\cos (A-B)+\sin C]$
$=1-2 \sin C[\cos (A-B)-\cos (A+B)]$
$=1-4 \sin A \sin B \sin C$
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