# Question #d92c6

Sep 22, 2017

Proving the identity will not fit in the answer, so look below.

#### Explanation:

So, we are trying to prove this: $\tan x - \csc x \sec x \left(1 - 2 {\cos}^{2} x\right) = \cot x$
Using this proof: ${\sin}^{2} x + {\cos}^{2} x = 1$...
$\tan x - \csc x \sec x \left({\sin}^{2} x + {\cos}^{2} x - 2 {\cos}^{2} x\right) = \cot x$
$\tan x - \csc x \sec x \left({\sin}^{2} x - {\cos}^{2} x\right) = \cot x$

Using these proofs: $\tan x = \sin \frac{x}{\cos} x$, $\csc x = \frac{1}{\sin} x$ and $\sec x = \frac{1}{\cos} x$.
$\sin \frac{x}{\cos} x - \frac{1}{\sin x \cos x} \left({\sin}^{2} x - {\cos}^{2} x\right) = \cot x$
$\sin \frac{x}{\cos} x - \frac{{\sin}^{2} x - {\cos}^{2} x}{\sin x \cos x} = \cot x$

Make both denominators the same on the left side only... (we leave the right side alone for proofs).
$\frac{{\sin}^{2} x - \left({\sin}^{2} x - {\cos}^{2} x\right)}{\sin x \cos x} = \cot x$
$\frac{{\sin}^{2} x - {\sin}^{2} x + {\cos}^{2} x}{\sin x \cos x} = \cot x$
$\frac{{\cos}^{2} x}{\sin x \cos x} = \cot x$
$\frac{\cos x}{\sin x} = \cot x$

Use this proof: $\cos \frac{x}{\sin} x = \cot x$
$\cot x = \cot x$