How do you prove #[(2tanx)/(1-tan^2x)]+[(1)/(2cos^2x-1)]= (cosx+sinx)/(cosx-sinx)#?

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Answer 1 · 2018-03-26T13:55:33

Please see below.

We know that,

#color(red)((1)tan2x=(2tanx)/(1-tan^2x)#

#color(red)((2)cos2x=2cos^2x-1=cos^2x-sin^2x#

#color(red)((3)sin2x=2sinxcosx,and cos^2x+sin^2x=1#

We have,

#(2tanx)/(1-tan^2x)+1/(2cos^2x-1)=(cosx+sinx)/(cosx-sinx)#

Now,

#LHS=(2tanx)/(1-tan^2x)+1/(2cos^2x-1#

#=tan2x+1/(cos2x).......to#Using #(1) and (2)#

#=(sin2x)/(cos2x)+1/(cos2x)#

#=(1+sin2x)/(cos2x)#

#=(sin^2x+cos^2x+2sinxcosx)/(cos^2x-sin^2x).....to#Using #(2) and (3)#

#=(cosx+sinx)^cancel(2)/(cancel((cosx+sinx))(cosx-sinx))#

#=(cosx+sinx)/(cosx-sinx)#

#=RHS#