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

May 14, 2015

We have to prove that $\frac{1 - \sin 2 x}{\cos 2 x} = \frac{\cos 2 x}{1 + \sin 2 x}$

To do this we transform left side:

$\frac{1 - \sin 2 x}{\cos 2 x} = \frac{\left({\sin}^{2} x + {\cos}^{2} x\right) - 2 \sin x \cos x}{\cos 2 x}$

$= \frac{{\sin}^{2} x - 2 \sin x \cos x + {\cos}^{2} x}{{\cos}^{2} x - {\sin}^{2} x}$

$= {\left(\sin x - \cos x\right)}^{2} / \left(\left(\cos x - \sin x\right) \left(\cos x + \sin x\right)\right)$

$= \frac{{\left(\sin x - \cos x\right)}^{2}}{- \left(\sin x - \cos x\right) \left(\sin x + \cos x\right)}$

$= \frac{\cos x - \sin x}{\cos x + \sin x}$

Now we expand the expresion by multiplying both numerator and denominator by $\left(\cos x + \sin x\right)$

So we get:

$\frac{\left(\cos x - \sin x\right) \left(\cos x + \sin x\right)}{{\left(\cos x + \sin x\right)}^{2}}$

$= \frac{{\cos}^{2} x - {\sin}^{2} x}{{\cos}^{2} x + 2 \cos x \sin x + {\sin}^{2} x}$

$= \frac{\cos 2 x}{1 + \sin 2 x}$