# How do you prove the following trig identity: tan(x + 45°) - tan(45° - x) ≡ 2tan2x?

Apr 15, 2015

Use $\tan \left(a + b\right) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$

and $\tan \left(a - b\right) = \frac{\tan a - \tan b}{1 + \tan a \tan b}$

and $\tan \left(2 a\right) = \tan \left(a + a\right) = \frac{2 \tan a}{1 - {\tan}^{2} a}$

And also use: $\tan \left(\frac{\pi}{4}\right) = 1$.

Other than that it's just algebra.

Once you get to

$\frac{\tan x + 1}{1 - \tan x} - \frac{1 - \tan x}{1 + \tan x}$

you'll want a common denominator, so you get

$\frac{{\left(\tan x + 1\right)}^{2} - {\left(1 - \tan x\right)}^{2}}{1 - {\tan}^{2} x}$

Now do the algebra to get:

$\frac{4 \tan x}{1 - {\tan}^{2} x} = 2 \left(\frac{2 \tan x}{1 - {\tan}^{2} x}\right) = 2 \tan 2 x$