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\frac { \sin \theta \sin 2 \theta + \sin 3 \theta \sin 60 } { \sin \theta \cos 2 \theta + \sin 3 \theta \cos 60 } = \tan 5 \theta |
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Answer» We will use the following formulae in this solution-- (1) Sin A . Sin B = (1/2) [ Cos (A-B) - Cos (A+B) ](2) Sin A Cos B = (1/2) [ Sin (A+B) + Sin (A-B) ] (3) Sin A + Sin B = 2 Sin (1/2) (A+B) . Cos (1/2)(A-B) (4) Sin A - Cos B = 2 Sin (1/2) (A-B) . Cos (1/2)(A+B) (5) Sin ( -x ) = - Sin x (6) Cos ( - x ) = Cos x Solution -- Numerator = (1/2) [ 2 SinA Sin 2A + 2 Sin 3A. Sin 6A ] --- Use formula (1) => (1/2) [ Cos (A-2A) - Cos (A+2A) + Cos (3A-6A) - Cos (3A+6A) ] => (1/2) [ Cos A - Cos 3A + Cos 3A - Cos 9A ] => (1/2) [ Cos A - Cos 9A ] .... Again use formula (1) => Sin 5A . Sin 4A Denominator = Sin A Cos 2A + Sin 3A.Cos 6A => (1/2) [ Sin (A+2A) + Sin (A-2A) + Sin ( 3A + 6A ) + Sin ( 3A - 6A ) ] => (1/2) [ Sin 3A - Sin A + Sin 9A - Sin 3A ] => (1/2) [ Sin 9A - Sin A ] => (1/2) [ 2 Cos 5A . Sin 4A ] => Cos 5A . Sin 4A Hence the Left Hand Side of the given identity = Numerator / Denominator => Sin 5A . Sin 4A / Cos 5A . Sin 4A => Tan 5A = RHS ………………. QED . I HOPE IT HELPS U. |
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