Question

Prove that sinA.sin2A+sin3A.sin6A/sinA.cosA+sin3A.cos6A=tan5A

Answer 1

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sin A sin 2A + sin 3A sin 6A/ sin A cos 2A + sin 3A cos 6A = tan 5A

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.