# Thread: Interesting identity

1. ## Interesting identity

Just got this in an exam:

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

Prove that $\displaystyle tanA.tanB.tanC = tanA + tanB + tanC$

I think there's an error though cause I find that they are not equal when I assign values to A, B, and C. Any help? I'm probably wrong

2. I think we're missing some information. I believe the identity is true only if A, B, and C are angles of the same triangle (A + B + C = 180°).

If so, break up the tangents into sines and cosines.

Then try using the product to sum formulas:

$\displaystyle \sin(\theta)\sin(\phi) = \frac{\cos(\theta - \phi) - \cos(\theta + \phi)}{2}$

$\displaystyle \cos(\theta)\cos(\phi) = \frac{\cos(\theta - \phi) + \cos(\theta + \phi)}{2}$

3. 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?

4. The product to sum formulas can be derived using those equations. For example:

$\displaystyle \cos(A+B) = \cos(A)\cos(B) - \sin(A)\sin(B)$ ---- 1
$\displaystyle \cos(A-B) = \cos(A)\cos(B) + \sin(A)\sin(B)$ ---- 2

1 + 2 gives:

$\displaystyle \cos(A+B) + \cos(A-B) = 2\cos(A)(cos(B)$ which is also

$\displaystyle \frac{\cos(A+B) + \cos(A-B)}{2} = \cos(A)cos(B)$

The product to sum formula for two sines is also derived in a similar way, adding $\displaystyle \sin(A+B)$ with $\displaystyle \sin(A-B)$

5. 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