(sin12A/sin4A)-(cos12A/cos4A)
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@=thita
we know that
Sin(A-B)=SinACosB-CosAsinB
Sin 2@=2Sin@cis@
2((Sin12A.Cos4A-Cos 12A.Sin4A)/(2Sin4A.cos4A))
2Sin(12A-4A)/(Sin8A)=2Sin8A/Sin8A
answer=2
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