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Question

sin2 42cos2 78=5+18


Solution

L.H.S.=sin2 42cos2 78=sin2 (9048)cos2 (9012)=cos2 48sin2 12=cos (48+12). cos(4812)[  cos (A+B). cos (AB)=cos2 Asin2 B]=cos 60.  cos 36=12.5+14    [  cos 36=5+14]=5+18
= RHS


Mathematics
RD Sharma
Standard XI

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