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\frac { \operatorname { sin } ( 4 A - 2 B ) + \operatorname { sin } ( 4 B - 2 A ) } { \operatorname { cos } ( 4 A - 2 B ) + \operatorname { cos } ( 4 B - 2 A ) } = \operatorname { tan } ( A + B ) |
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Answer» L.H.S.=(2sin(A+B)cos(3A-3B))/(2cos(A+B)cos(3A-3B)) =sin(A+B)/cos(A+B) =tan(A+B) =R.H.S.Hence Proved |
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