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)`