1. ## trig identities

Can someone help me with this identity, I can't seem to get it,

tan2x=2tanx/1-tan^2x

thanks.

2. Hello, John!

I must assume that we are allowed some other identities:

. . $\sin2\theta \:=\:2\sin\theta\cos\theta \qquad\quad \cos2\theta \:=\: \cos^2\!\theta - \sin^2\!$

$\tan2x\:=\:\frac{2\tan x}{1-\tan^2\!x}$

We have: . $\tan2x \:=\:\frac{\sin2x}{\cos2x} \:=\:\frac{2\sin x\cos x}{\cos^2\!x - \sin^2\!x}$

$\text{Divide top and bottom by }\cos^2\!x\!:\quad \frac{\dfrac{2\sin x\cos x}{\cos^2\!x}} {\dfrac{\cos^2\!x}{\cos^2\!x} - \dfrac{\sin^2\!x}{\cos^2\!x}}$

. . $\text{and we have: }\;\frac{2\dfrac{\sin x}{\cos x}} {1 - \left(\dfrac{\sin x}{\cos x}\right)^2} \;=\;\frac{2\tan x}{1 - \tan^2\!x}$