Question

Sin 2A + Sin 2 B + Sin 2 (A-B) = 4 Sin A. Cos B. Cos (A-B)
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Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

On multiplying and dividing LHS by 2, we get 2(sin^2(A) -sin^2(B))/ 2(sin(A) cos(A) -sin(B) cos(B)), on applying 2sin^2(x)= 1–cos2x formula to numerator and sin2x=2sinxcosx to denominator we get, (cos2B-cos2A)/(sin2A-sin2B), now to numerator apply cosC- cosD=2sin((C+D)/2)*sin((D-C)/2) and to denominator apply formula sinC-sinD=2cos((C+D)/2)*sin((C-D)/2), we get sin(A+B)sin(A-B)/(cos(A+B)sin(A-B)), on solving this we get tan(A+B)