# How do you simplify (tan 4Θ + tan 2Θ) / (1-tan 4Θtan 2Θ)?

Oct 9, 2016

$\frac{\tan 4 \theta + \tan 2 \theta}{1 - \tan 4 \theta \tan 2 \theta} = \tan 6 \theta$

#### Explanation:

The sum of angles formulae for $\sin$ and $\cos$ can be written:

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \sin \beta \cos \alpha$

$\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$

Hence we find:

$\tan \left(\alpha + \beta\right) = \frac{\sin \left(\alpha + \beta\right)}{\cos \left(\alpha + \beta\right)}$

$\textcolor{w h i t e}{\tan \left(\alpha + \beta\right)} = \frac{\sin \alpha \cos \beta + \sin \beta \cos \alpha}{\cos \alpha \cos \beta - \sin \alpha \sin \beta}$

$\textcolor{w h i t e}{\tan \left(\alpha + \beta\right)} = \frac{\left(\sin \alpha \cos \beta + \sin \beta \cos \alpha\right) \div \left(\cos \alpha \cos \beta\right)}{\left(\cos \alpha \cos \beta - \sin \alpha \sin \beta\right) \div \left(\cos \alpha \cos \beta\right)}$

$\textcolor{w h i t e}{\tan \left(\alpha + \beta\right)} = \frac{\sin \frac{\alpha}{\cos} \alpha + \sin \frac{\beta}{\cos} \beta}{1 - \sin \frac{\alpha}{\cos} \alpha \sin \frac{\beta}{\cos} \beta}$

$\textcolor{w h i t e}{\tan \left(\alpha + \beta\right)} = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

So the sum of angles formula for $\tan$ may be written:

$\tan \left(\alpha + \beta\right) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$

Putting $\alpha = 4 \theta$ and $\beta = 2 \theta$, we find:

$\frac{\tan 4 \theta + \tan 2 \theta}{1 - \tan 4 \theta \tan 2 \theta} = \tan \left(4 \theta + 2 \theta\right) = \tan 6 \theta$