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*tanx##color(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

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