Product of Trigonometric Ratios in Terms of Their Sum
2 sin2 β+4 co...
Question
2sin2β+4cos(α+β)sinαsinβ+cos2(α+β)=
A
sin2α
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B
cos2β
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C
cos2α
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D
sin2β
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Solution
The correct option is Ccos2α cos2(α+β)=2cos2(α+β)−1,2sin2β=1−cos2β
L.H.S. =−cos2β+2cos(α+β)[2sinαsinβ+cos(α+β)] =−cos2β+2cos(α+β)cos(α+β) =−cos2β+(cos2α+cos2β)=cos2α.