Question

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

Answer 1

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)`

Answer 2

`(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