Hello MarcoMP Originally Posted by

**MarcoMP** Ok I have checked the information and I left out the first line and you were correct, it only applies to a triangle.

On the level that we do trigenometry, we can do compound angles $\displaystyle sin(A+B) = sinA.CosB + CosA.SinB$

Then changing from $\displaystyle Cos2A = 2Cos^2A -1 = 1 -2Sin^2A = Cos^2A - Sin^2A$

And from $\displaystyle Sin2A = 2SinA.CosA$

Do you perhaps know how i would go about proving this identity using these rules?

Starting with the formula you initially quoted, then:

$\displaystyle \tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)}$

$\displaystyle =\frac{\sin A \cos B+\cos A \sin B}{\cos A\cos B -\sin A \sin B}$

$\displaystyle =\frac{\tan A + \tan B}{1-\tan A \tan B}$, by dividing top-and-bottom by $\displaystyle \cos A \cos B$

$\displaystyle \Rightarrow \tan(A+B) - \tan(A+B)\tan A\tan B=\tan A +\tan B$

But if $\displaystyle A+B= 180 - C$, then $\displaystyle \tan(A+B) = -\tan C.$

$\displaystyle \Rightarrow -\tan C +\tan A\tan B\tan C=\tan A + \tan B$

$\displaystyle \Rightarrow \tan A\tan B\tan C =\tan A+\tan B+\tan C$

Grandad