Tan2x formula in terms of tanx is as follows: $\dfrac{2\tan x}{1-\tan^2x}$. Here, we will also derive the tan2x formula in terms of sinx and cosx along with some examples. These formulas are very useful to solve trigonometric equations and simplify trigonometric expressions.

## Tan2x Formula

Tan2x is the tangent function of a double angle 2x. The formula of tan2x in terms of tanx is given as follows.

$\tan 2x=\dfrac{2\tan x}{1-\tan^2x}$.

## Tan2x in Terms of Tanx

To derive the tan 2x formula, we will use the trigonometric formula below:

$\tan(a+b)$ $=\dfrac{\tan a +\tan b}{1-\tan a\tan b}$

At first, we will write 2x as x+x, and then we will apply the above formula. By doing so, we get that

$\tan 2x=\tan(x+x)$

$=\dfrac{\tan x+\tan x}{1-\tan x \tan x}$

$=\dfrac{2\tan x}{1-\tan^2x}$

## Tan2x Formula Proof

Now, we will prove the tan2x formula using sin2x and cos2x formulas which are given below.

- $\tan x=\dfrac{\sin x}{\cos x}$
- $\sin 2x =2\sin x\cos x$
- $\cos 2x=\cos^2x-\sin^2x$

Then we have $\tan 2x =\dfrac{\sin 2x}{\cos 2x}$

$=\dfrac{2\sin x \cos x}{\cos^2x-\sin^2x}$

Dividing both the numerator and denominator by $\cos^2x$, we get that

$\tan 2x=\dfrac{\dfrac{2\sin x \cos x}{\cos^2x}}{\dfrac{\cos^2x-\sin^2x}{\cos^2x}}$

$=\dfrac{2\dfrac{\sin x}{\cos x} cdot \dfrac{\cos x}{\cos x}}{\dfrac{\cos^2x}{\cos^2x}-\dfrac{\sin^2x}{\cos^2x}}$

$=\dfrac{2\tan x}{1-\tan^2x}$

Thus, we have proved the formula of tan2x which is $\dfrac{2\tan x}{1-\tan^2x}$.

## Tan2x in Terms of Sinx

Now we will derive the tan2x formula, in terms of sinx. In order to do so, we will go through the following way:

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$=\dfrac{2\sin x \cos x}{1-2\sin^2x}$ as cos2x=1-2sin^{2}x

$=\dfrac{2\sin x}{1-2\sin^2x} \sqrt{1-\sin^2x}$ as cos^{2}x=1-sin^{2}x

Tan2x in Terms of Cosx

Now we will derive the tan2x formula, in terms of cosx. In order to do so, we will go through the following way:

$\tan 2x=\dfrac{\sin 2x}{\cos 2x}$

$=\dfrac{2\sin x \cos x}{2\cos^2x-1}$ as $\cos 2x=2\cos^2x-1$

$=\dfrac{2\cos x}{2\cos^2x-1} \sqrt{1-\cos^2x}$ as $\sin^2x=1-\cos^2x$

**Also Read: **

**sin3x Formula in terms of sinx**

**Question:** Find the Value of $\tan 120^\circ$.

*Answer: *

We have from the above that $\tan 2x=\dfrac{2\tan x}{1-\tan^2x}$.

Put $x=60^\circ$

Then $\tan 120^\circ=\dfrac{2\tan 60^\circ}{1-\tan^2 60^\circ}$

$=\dfrac{2\tan 60^\circ}{1-\tan^2 60^\circ}$ as $\tan 60^\circ=\sqrt{3}$

$=\dfrac{2\sqrt{3}}{1-(\sqrt{3})^2}$

$=\dfrac{2\sqrt{3}}{1-3}$

$=-\sqrt{3}$.

So the value of tan120 degrees is equal to root 3.

## FAQs

**Q1: What is the tan2x formula in terms of tanx?**

**Answer:** The tan2x formula in terms of tanx is given as follows: tan2x=2tanx/(1-tan^{2}x).

**Q2: What is the tan2x formula in terms of sinx?**

**Answer:** The tan2x formula in terms of sinx is given as follows: tan2x $=\dfrac{2\cos x}{2\cos^2x-1} \sqrt{1-\cos^2x}$.