# Math Help - Addition Formulae (iii) trigonometry identity

1. ## Addition Formulae (iii) trigonometry identity

OK, after help from here, I tried 4 more questions and thought i got the hang of this. Now i have another dilemma. :S

By expressing: tan(alpha + beta) = sin(alpha + beta) / cos(alpha + beta)

Prove that: tan(alpha + beta) = tan alpha + tan beta / 1 - tan alpha tan beta

Here are my first steps:
= sin(alpha + beta) / cos(alpha + beta)
= sin alpha cos beta + cos alpha sin beta / cos alpha cos beta - sin alpha sin beta

now im stuck,
should i cancel out the cos and sin from top and bottom?
i tried that, but that didn't work. after i thought, since i was lookin for tan in my answer, i would sin / cos = tan
that also did not work out.... (or maybe i did something wrong)

2. You have,
$\tan (x+y)=\frac{\sin(x+y)}{\cos(x+y)}$
$\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$
Divide through when non-zero by $\cos x\cos y$ the numerator and denominator,
$\frac{\tan x+\tan y}{1-\tan x\tan y}$

3. Divide through when non-zero by $\cos x\cos y$ the numerator and denominator,

^

youv lost me there, can u explain clearer,
what do you mean by "Divide through when non-zero"

4. Originally Posted by r_maths
youv lost me there, can u explain clearer,
$\frac{\sin x\cos y+\cos x\sin y}{\cos x\cos y-\sin x\sin y}$
Divide through,
$\frac{\frac{\sin x \cos y}{\cos x\cos y}+\frac{\cos x\sin y}{\cos x\cos y}}{\frac{\cos x\cos y}{\cos x\cos y}-\frac{\sin x\sin y}{\cos x\cos y}}$
Thus, you are going to get some canceling.
And you also need to use the fact that,
$\frac{\sin x}{\cos x}=\tan x$
what do you mean by "Divide through when non-zero"
I am not saying if the cosines are not zero you can divide through, thus the formula for tangent addition will hold true.

5. thanks,