How do you prove (1+tan^2x)/(1-tan^2x)=1+tan2x.tanx ?

2 Answers
Yahia M.
May 1, 2018

Check the explanation below

Explanation:

(1+tan^2x)/(1-tan^2x)

=(1color(green)(-tan^2x+tan^2x)+tan^2x)/(1-tan^2x)

=(1-tan^2x)/(1-tan^2x)+(2tan^2x)/(1-tan^2x)

color(red)(tanx=sinx/cosx

=1+(2color(green)(sin^2x/cos^2x))/(1-color(green)(sin^2x/cos^2x)

=1+(2sin^2x)/(cos^2x-sin^2x)

Using the Double Angle Identities

  • color(red)(cos^2x-sin^2x=cos2x)
  • color(red)(2sinxcosx=sin2x

=1+(2sin^2x)/(cos2x)

=1+2sin^2x/(cos2x)color(green)(xxcosx/cosx

=1+(sinx*(2sinxcosx))/(cosx*cos2x)

=1+color(green)(sinx*color(blue)(sin2x))/(color(green)(cosx*color(blue)(cos2x)

=1+tan2x*tanxcolor(green)(rarr " R.T.P"

I hope this was helpful :)

Steve M
May 1, 2018

We seek to prove the identity:

(1+tan^2x)/(1-tan^2x) -=1+tan2xtanx

We can utilise the tangent double angle formula:

tan2A -= (2tanA)/(1-tan^2A)

Consider the RHS:

RHS = 1+tan2xtanx

\ \ \ \ \ \ \ \ = 1+((2tanx)/(1-tan^2x))tanx

\ \ \ \ \ \ \ \ = 1+(2tan^2x)/(1-tan^2x)

\ \ \ \ \ \ \ \ = ((1-tan^2x)+(2tan^2x))/(1-tan^2x)

\ \ \ \ \ \ \ \ = (1+tan^2x)/(1-tan^2x)

\ \ \ \ \ \ \ \ = LHS \ \ \ QED

Related questions
See all questions in Proving Identities
Impact of this question
5819 views around the world
You can reuse this answer
Creative Commons License