Question

property of
tan(A + B)

Answer 1

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I am not actually sure what do you exactly mean by proving it but, I will try.

To prove it, first we need to know the basic formula of sin(A+B) and cos(A+B).

Sin(A+B)= sinA. CosB+ CosA.SinB--(1)
Cos(A+B)= cosA. CosB-sinA.sinB--(2)

As we know, that sinA/CosA= TanA
The same way,  if we divide eqn (1) by (2)
We get,
sin(A+B)/Cos(A+B) = sinA.CosB+CosA.SinB / cosA. CosB- sinA. sinB
 
Now we have,
Tan(A+B) =sinA.CosB+CosA.SinB / cosA. CosB- sinA. sinB

And then dividing, the numerator and denominator of RHS by cosA.cosB

Answer 2

complete geometric derivation of the formula for tan(A + B) is complicated. An easy way is to derive it from the two formulas that you have already done. In any angle, the tangent is equal to the sine divided by the cosine. Using that fact, tan(A + B) = sin(A + B)/cos(A + B). In a way that does it, but you can expand that to: 
tan(A+B)=sin Acos B+cos A sin Bcos Acos B−sin A sin Btan⁡(A+B)=sin⁡ Acos⁡ B+cos⁡ A sin⁡ Bcos⁡ Acos⁡ B−sin⁡ A sin⁡ B 
Divide through top and bottom by cos A cos B, which turns all the terms into tangents, giving: 
tan(A+B)=tan A+tan B1−tan A tan Btan⁡(A+B)=tan⁡ A+tan⁡ B1−tan⁡ A tan⁡ B