# prove tan (alpha + beta) =( tan alpha + tan beta) / (1 - tan alpha tan beta)

tan (α + β) = (tanα + tanβ) / (1 - tanα tanβ)

Now take ' tanα + tanβ '

[ Recall: tanA = sinA / cosA ]

tanα + tanβ = (sinα / cosα) + (sinβ / cosβ)

Here LCM of cosα, cosβ is cosαcosβ

tanα + tanβ = (sinαcosβ + sinβcosα ) / (cosαcosβ) ------------------> (1)

And  1 - tanα tanβ = 1 - (sinα/cosα)(sinβ/cosβ)

1 - tanα tanβ = 1 - (sinαsinβ) / (cosαcosβ)

Here LCM of 1, cosαcosβ is cosαcosβ

1 - tanα tanβ = [cosαcosβ - sinαsinβ] / (cosαcosβ) --------------> (2)

Now solve (1) / (2)

[tanα + tanβ]/[1 - tanα tanβ] = [(sinαcosβ + sinβcosα )/(cosαcosβ)] / [cosαcosβ - sinαsinβ]/(cosαcosβ)]

Simplify

[tanα + tanβ] / [1 - tanα tanβ] = [(sinαcosβ + sinβcosα )] / [ cosαcosβ - sinαsinβ]

sine of sum is sin (α + β) = sinα cosβ + cosαsinβ

cosine of sum is cos (α + β) = cosα cosβ - sinαsinβ

So,

[tanα + tanβ] / [1 - tanα tanβ] = sin (α + β) / cos (α + β)

Recall: sinA / cosA = tan A

[tanα + tanβ] / [1 - tanα tanβ] = tan (α + β)

There fore

tan (α + β) = (tanα + tanβ) / (1 - tanα tanβ)