Question

1) Prove that: cos 3x-cos 7x / sin 9x+sin x =cos 2x.tan 4x-sin 2x​

Answer 1

To prove:

cos 3x-cos 7x / sin 9x+sin x =cos 2x.tan 4x-sin 2x​

Formula used:

sinC+sinD =2sin(C+D/2)cos(C-D/2)

CosC-CosD=2sin(C+D/2)sin(D-C/2)

sin(A -B) = sinAcosB - cosAsinB

Proof:

L.H.S.,

cos 3x-cos 7x / sin 9x+sin x = $\frac{2sin5xsin2x}{2sin5xcos4x}$

                                                =  $\frac{sin2x}{cos4x}$

R.H.S.,

cos 2x.tan 4x-sin 2x​ = $\frac{cos2x . sin4x}{cos4x}$ - $sin2x$

                                 = $\frac{cos2x . sin4x- sin2xcos4x}{cos4x}$

                                 = $\frac{sin(4x-2x)}{cos4x}$

                                 = $\frac{sin2x}{cos4x}$

L.H.S = R.H.S.

Hence, cos 3x-cos 7x / sin 9x+sin x =cos 2x.tan 4x-sin 2x​

Proved

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