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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!

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    $\begingroup$ Welcome to StackExchange! What have you tried? Have you tried using the double angle formula for tangent? $\endgroup$
    – rb612
    Jun 22, 2017 at 2:25
  • $\begingroup$ No afford, and people giving their solutions, let him try! $\endgroup$
    – MAN-MADE
    Jun 22, 2017 at 2:44

4 Answers 4

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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.

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

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  • $\begingroup$ That makes it very clear; thanks! $\endgroup$ Jun 22, 2017 at 3:14
  • $\begingroup$ @Hamza Qayyum Welcome! $\endgroup$ Jun 22, 2017 at 3:16
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For fun, here's a trigonograph:

enter image description here

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

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

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