How do you prove #cos2x= (1-tan^2(x))/ (1+tan^2(x))#?

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Answer 1 · 2015-04-16T17:03:51

In this way:

(remembering that #tanx=sinx/cosx# and #sin^2x+cos^2x=1#),

the second member becomes:

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

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

#=cos^2x-sin^2x#,

that is the development of the formula of #cos2x#.