Proving trig identity $\tan(2x)−\tan(x)=\frac{\tan(x)}{\cos(2x)}$ I'm currently stumped on proving the trig identity below:
$\tan(2x)-\tan (x)=\frac{\tan (x)}{\cos(2x)}$
Or, alternatively written as:
$\tan(2x)-\tan (x)=\tan (x)\sec(2x)$
Help on deriving it would be appreciated; thanks!
 A: Using some trig-identities we have:
$$\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$$
and
$$\cos(2x)=2\cos^2(x)-1$$
and
$$\tan^2(x)=\sec^2(x)-1$$
we have (on the left hand side):
$$\begin{align}\tan(2x)-\tan(x)&=\frac{2\tan(x)}{1-\tan^2(x)}-\tan(x)\\&=\frac{2\tan(x)-\tan(x)+\tan^3(x)}{1-\tan^2(x)}\\&=\tan(x)\frac{1+\tan^2(x)}{1-\tan^2(x)}\\&=\tan(x)\frac{1+\sec^2(x)-1}{1-\sec^2(x)+1}\\&=\tan(x)\left[\frac{2-\sec^2(x)}{\sec^2(x)}\right]^{-1}\\&=\tan(x)\left[2\cos^2(x)-1\right]^{-1}\\&=\tan(x)[\cos(2x)]^{-1}\\&=\tan(x)\sec(2x)\end{align}$$
and we are done.
A: $$\tan2x-\tan{x}=\frac{\sin2x}{\cos2x}-\frac{\sin{x}}{\cos{x}}=$$
$$=\frac{\sin2x\cos{x}-\cos{2x}\sin{x}}{\cos2x\cos{x}}=$$
$$=\frac{\sin{x}}{\cos{x}\cos2x}=\frac{\tan{x}}{\cos{2x}}$$
A: For fun, here's a trigonograph:

$$\tan 2\theta = \tan\theta + \tan\theta \sec 2\theta$$
A: hint
$$\cos (2x)=\frac {1-\tan^2 (x)}{1+\tan^2 (x)} $$
$$\sin(2x)=\frac {2\tan (x)}{1+\tan^2 (x)} $$
thus
$$\sin(2x)-\tan (x)\cos (2x)=$$
$$\frac {2\tan {x}-\tan (x)+\tan^3 (x)}{1+\tan^2 (x)}$$
$$=\tan (x) $$ Done.
