# Show that (2cos^2(x/2)tanx)/(tanx) = (tanx+cosxtanx)/(tanx)?

Oct 24, 2015

Okay, let's consider dividing both sides by $\tan x$ to begin with:

$\frac{2 {\cos}^{2} \left(\frac{x}{2}\right) \cancel{\tan x}}{\cancel{\tan x}} = \frac{\cancel{\tan x} \left(1 + \cos x\right)}{\cancel{\tan x}}$

$2 {\cos}^{2} \left(\frac{x}{2}\right) = 1 + \cos x \implies 2 {\cos}^{2} \left(\frac{x}{2}\right) - 1 = \cos x$

Notice how we can use the identity ${\sin}^{2} \left(u\right) + {\cos}^{2} \left(u\right) = 1$, where $u = \frac{x}{2}$:

$2 {\cos}^{2} \left(\frac{x}{2}\right) - \left[{\sin}^{2} \left(\frac{x}{2}\right) + {\cos}^{2} \left(\frac{x}{2}\right)\right] = \cos x$

$2 {\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right) - {\cos}^{2} \left(\frac{x}{2}\right) = \cos x$

${\cos}^{2} \left(\frac{x}{2}\right) - {\sin}^{2} \left(\frac{x}{2}\right) = \cos x$

Then, we can use the identity $\cos \left(u + v\right) = \cos u \cos v - \sin u \sin v$, where $u = v = \frac{x}{2}$ and work backwards:

$\cos \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right) - \sin \left(\frac{x}{2}\right) \sin \left(\frac{x}{2}\right) = \cos x$

$\cos \left(\frac{x}{2} + \frac{x}{2}\right) = \cos x$

$\textcolor{b l u e}{\cos x = \cos x}$