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

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Answer 1 · 2015-04-27T03:08:52

Multiply the left side in the numerator and denominator by 1-tanx and simplify to get the answer:

$(1-tanx)^2 / (1-tan^2 x)$

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

= $(sec^2x -2tanx)/((cos^2x -sin^2x)/cos^2x)$ (because $1+tan^2x = sec^2x$ )

= $(1-2tan x cos^2x)/cos(2x)$ (because $cos^2x- sin^2x=cos 2x$ )

= $(1-sin2x)/cos(2x)$

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