# How do you show that sec(2x) + tan(2x) + 1 = 2/(1 - tanx)?

Nov 26, 2016

First of all, $\sec \theta = \frac{1}{\cos} \theta$ and $\tan \theta = \sin \frac{\theta}{\cos} \theta$.

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

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

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

We now establish the double angle identities. I won't go into the proofs for these, but I would just like to emphasize that these formulas are very important.

$\sin 2 \theta = 2 \sin \theta \cos \theta$
$\cos 2 \theta = 1 - 2 {\sin}^{2} \theta = 2 {\cos}^{2} \theta - 1 = {\cos}^{2} \theta - {\sin}^{2} \theta$

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

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

Factor:

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

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

Identity proved!

Hopefully this helps!