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Question : 23 of 160
Marks:
+1,
-0
Solution:
A+B+C=‌ cos‌2‌A+cos‌2‌B+cos‌2‌C =2‌cos(‌)⋅cos(‌)+cos‌2‌C =2‌cos(‌−C)⋅cos(A−B)+cos‌2‌C =2(−sin‌C)⋅cos(A−B)+(−2sin‌2C+1) =−2sin‌C[cos(A−B)+sin‌C]+1 =−2sin‌C[cos(A−B)+sin‌(‌−(A+B))]+1 =−2sin‌C[cos(A−B)−cos(A+B)]+1 =−2sin‌C(2sin‌A⋅sin‌B)+1 =−4sin‌A⋅sin‌B⋅sin‌C+1 ⇒4sin‌A⋅sin‌B⋅sin‌C+(cos‌2‌A+cos‌2‌B+cos‌2‌C)=1=−sin‌(A+B+C)[∵sin‌(A+B+C)=sin‌‌=−1]
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