How do you prove #(1 + tanx) / (1 - tanx) = (1 + sin2x) / (cos2x)#?

2 Answers
Mar 19, 2016

his is an identity. For proof, please see below.

Explanation:

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

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

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

Now multiplying numerator and denominator by #(cosx+sinx)#, we get

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

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

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

Mar 19, 2016

#(1+tanx)/(1-tanx)#
multiplying both numerator and denominator by #(1+tanx)# we have
LHS#=(1+tanx)/(1-tanx)xx(1+tanx)/(1+anx)=(1+tanx)^2/(1-tan^2x)=(1+tan^2+2tanx)/(1-tan^2x)#
dividing both numerator and denominator by #(1+tan^2x)# we have

#((1+tan^2x)/(1+tan^2x)+(2tanx)/(1+tan^2x))/((1-tan^2x)/(1+tan^2x))#
#(1+sin2x)/ (cos2x)#= RHS