Prove that #(1-2sin^2x)/(1+sin2x) = (1-tanx)/(1+tanx)#?

Question

6895 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-10T15:07:25

Please see below.

To prove #(1-2sin^2x)/(1+sin2x) = (1-tanx)/(1+tanx)#, let us start from left hand side.

#(1-2sin^2x)/(1+sin2x)# is equivalent to

#(1-sin^2x-sin^2x)/(cos^2x+sin^2x+2sinxcosx)#

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

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

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

Now dividing numerator and denominator by #cosx#

= #(1-sinx/cosx)/(1+sinx/cosx)#

= #(1-tanx)/(1+tanx)#