Question

2 sin 3x . cos 2x - 2sin 3x​

Answer 1

Answer:

FOLLOW ME

(sin5x-2sin3x+sinx)/(cos5x-cosx)=tanx

LHS

=[(sin5x-sin3x)-(sin3x-sinx)]/(cos5x-cosx)

=[(2cos4x.sinx)-(2cos2x.sinx)]/(-2sin3x.sin2x)

=2sinx.(cos4x-cos2x)/(-2sin3x.2sinx.cosx)

=(cos4x-cos2x)/(-2sin3x.cosx)

=(-2sin3x.sinx)/(-2sin3x,cosx)

=(sinx)/(cosx)

=tanx , Proved.

SECOND -METHOD:-

LHS.

=(sin5x+sinx-2sin3x)/(cos5x-cosx)

=(2sin3x.cos2x-2sin3x)/(-2sin3x.sin2x)

=2sin3x(cos2x-1)(-2sin3x.sin2x)

=(cos2x-1)/(-sin2x)

=(1–2sin^2x-1)/(-2sinx.cosx)

÷(-2sin^2x)/(-2sinx.cosx)

=sinx/cosx

=tanx , Proved.