# How do you simplify tan(x+y) to trigonometric functions of x and y?

Jan 6, 2016

$\tan \left(x + y\right) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$

#### Explanation:

This can be expanded through the tangent angle addition formula:

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

Thus,

$\tan \left(x + y\right) = \frac{\tan x + \tan y}{1 - \tan x \tan y}$

The tangent addition formula can be found using the sine and cosine angle addition formulas.

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

Since $\tan x = \sin \frac{x}{\cos} x$,

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

This can be written in terms of tangent by dividing both the numerator and denominator by $\cos \alpha \cos \beta$.

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

Final round of simplification yields:

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