How do you prove $1 + \frac{{tan}^{2} x}{1 - {tan}^{2} x} = \frac{1}{{cos}^{2} x - {sin}^{2} x}$ $1 + tan^2x/(1-tan^2x) = 1/(cos^2x - sin^2x)$ ?

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Answer 1 · 2015-12-07T18:46:47

That's true only if $cos^2 x = 1 Rightarrow x in pi ZZ$

$1 + tan^2 x / (1- tan^2 x) = 1 + frac{sin^2 x/ cos^2 x}{ 1- sin^2 x/cos^2 x} =$

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

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

How do you prove 1 + tan^2x/(1-tan^2x) = 1/(cos^2x - sin^2x) 1+tan2x1−tan2x=1cos2x−sin2x 1 + tan^2x/(1-tan^2x) = 1/(cos^2x - sin^2x) ?