Recall that

`tan\ theta=(sin\ theta)/(cos\ theta)`

So, letting θ = α + β, and expanding using our new sine and cosine identities, we have:

`tan(alpha+beta)=(sin(alpha+beta))/(cos(alpha+beta))=(sin alpha cos beta+cos alpha sin beta)/(cos alpha cos beta-sin alpha sin beta)`

Dividing numerator and denominator by cos α cos β:

`=(sin\ alpha\ cos\ beta+cos\ alpha\ sin\ beta)/(cos\ alpha\ cos\ beta-sin\ alpha\ sin\ beta)-:(cos\ alpha\ cos\ beta)/(cos\ alpha\ cos\ beta`

Simplifying gives us:

`tan(alpha+beta)=(tan\ alpha+tan\ beta)/(1-tan\ alpha\ tan\ beta)`

Replacing β with (-β) gives us

`tan(alpha-beta)=(tan\ alpha-tan\ beta)/(1+tan\ alpha\ tan\ beta)`

[The tangent function is odd, so tan(-β) = - tan β]

We have proved the two tangent of the sum and difference of two angles identities:

`tan(alpha+beta)=(tan\ alpha+tan\ beta)/(1-tan\ alpha\ tan\ beta)`

`tan(alpha-beta)=(tan\ alpha-tan\ beta)/(1+tan\ alpha\ tan\ beta)`