need help on a question on the + and - of angles (1 Viewer)

[frippyful]

hi, i have no idea how to do this question, i have the answer, but i guess there is something about trig signs that i dont know of, like can you divide them, because i got close to the answer but i still had a random tan there where the answer didnt. i need to prove this identity:

tan(x + pi/6) = root 3 x tanx + 1 / root 3 - tanx

so can someone please walk me through the process especially with tan. and sorry about the representation of the question, i dont know how else to get it on here, so maybe carefully go through it and write it down first

thanks

 

[deswa1]

This question involves multiple angles with tan:

<a href="http://www.codecogs.com/eqnedit.php?latex=tan(x@plus;y)=\frac{tanx@plus;tany}{1-tanxtany}\\ tan(x@plus;\frac{\pi}{6})=\frac{tanx@plus;tan(\frac{\pi}{6})}{1-tanxtan(\frac{\pi}{6})} \\ tan(\frac{\pi}{6})=\frac{1}{\sqrt3} \\ \therefore tan(x@plus;\frac{\pi}{6})=\frac{tanx@plus;\frac{1}{\sqrt3}}{1-\frac{tanx}{\sqrt3}}\\ =\frac{\sqrt3tanx@plus;1}{\sqrt3-tanx}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?tan(x+y)=\frac{tanx+tany}{1-tanxtany}\\ tan(x+\frac{\pi}{6})=\frac{tanx+tan(\frac{\pi}{6})}{1-tanxtan(\frac{\pi}{6})} \\ tan(\frac{\pi}{6})=\frac{1}{\sqrt3} \\ \therefore tan(x+\frac{\pi}{6})=\frac{tanx+\frac{1}{\sqrt3}}{1-\frac{tanx}{\sqrt3}}\\ =\frac{\sqrt3tanx+1}{\sqrt3-tanx}" title="tan(x+y)=\frac{tanx+tany}{1-tanxtany}\\ tan(x+\frac{\pi}{6})=\frac{tanx+tan(\frac{\pi}{6})}{1-tanxtan(\frac{\pi}{6})} \\ tan(\frac{\pi}{6})=\frac{1}{\sqrt3} \\ \therefore tan(x+\frac{\pi}{6})=\frac{tanx+\frac{1}{\sqrt3}}{1-\frac{tanx}{\sqrt3}}\\ =\frac{\sqrt3tanx+1}{\sqrt3-tanx}" /></a>