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!

• Welcome to StackExchange! What have you tried? Have you tried using the double angle formula for tangent? Jun 22, 2017 at 2:25
• No afford, and people giving their solutions, let him try! Jun 22, 2017 at 2:44

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.

$$\tan2x-\tan{x}=\frac{\sin2x}{\cos2x}-\frac{\sin{x}}{\cos{x}}=$$ $$=\frac{\sin2x\cos{x}-\cos{2x}\sin{x}}{\cos2x\cos{x}}=$$ $$=\frac{\sin{x}}{\cos{x}\cos2x}=\frac{\tan{x}}{\cos{2x}}$$

• That makes it very clear; thanks! Jun 22, 2017 at 3:14
• @Hamza Qayyum Welcome! Jun 22, 2017 at 3:16

For fun, here's a trigonograph:

$$\tan 2\theta = \tan\theta + \tan\theta \sec 2\theta$$

• Wow that's really cool. Awesome tool for visualising the identity. Thanks! Jun 22, 2017 at 4:02

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.