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

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

$cos x = \frac{1 - {(\frac{1 - cos x}{sin x})}^{2}}{1 + {(\frac{1 - cos x}{sin x})}^{2}}$ $Cosx=(1-((1-cosx)/sinx)^2)/(1+((1-cosx)/sinx)^2)$

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

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

$cos x = \frac{\frac{2 cos x - 2 {cos}^{2} x}{{sin}^{2} x}}{\frac{2 - 2 cos x}{{sin}^{2} x}}$ $Cosx=((2cosx-2cos^2x)/sin^2x)/((2-2cosx)/sin^2x)$

$Cosx=(cancel(2)cosxcancel((1-cosx)))/cancel(sin^2x)*cancel(sin^2x)/(cancel2cancel((1-cosx))$

$cosx=cosx$