Answer:
Option C
Explanation:
Given , A+B+C =270°
and $\cos 2A+\cos 2B+\cos 2C+4\sin A \sin B \sin C$
= $2 \cos (A+B) \cos (A-B)+1-2 \sin ^{2} C + 4 \sin A\sin B \sin C $
=$2 \cos (270 ^{0}-C) \cos (A-B)+1-2 \sin^{2} C+ 4 \sin A \sin B \sin C$
= $ 1-2 \sin C [ \cos (A-B)+\sin C]+ 4 \sin A \sin B \sin C$
= $1-2 \sin C[ \cos (A-B)+\sin C] +4 \sin A \sin B \sin C $
=$1-2 \sin C[ \cos (A-B)- \cos (A+B) ]+4 \sin A \sin B \sin C$
=$ 1-4 \sin A \sin B \sin C+4 \sin A \sin B \sin C=1$
