Question #14f11

1 Answer
Feb 19, 2018

Please see below.

Explanation:

We will use #cos2x=1-2sin^2x# and #sin2x=2sinx*cosx#.

#LHS=(1-cos2x-sinx)/(sin2x-cosx)#

#=(1-(1-2sin^2x)-sinx)/(2sinxcosx-cosx)#

#=(2sin^2x-sinx)/(2sinxcosx-cosx)#

#=(sinx*(2sinx-1))/(cosx(2sinx-1)#

#=tanx=RHS#