1.

\operatorname { sin } ( A - B ) ( \operatorname { tan } A + \operatorname { tan } B ) = \operatorname { sin } ( A + B ) ( \operatorname { tan } A - \operatorname { tan } B

Answer»

Sin(A-B) (TanA+TanB) =Sin(A+B) (TanA-TanB) Taking L. H. S. first=(SinA-SinB) (SinA/CosA+SinB/CosB) =Sin^2A/CosA-Sin^2B/CosBNow taking R. H. S.... =(SinA+SinB) (SinA/CosA-SinB/CosB) =Sin^2A/CosA-Sin^2B/CosBHence.. L. H. S=R. H. S. proved

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