# Verify? (cotx-tanx)/cosxsinx=csc^2x-sec^2x

See below:

#### Explanation:

$\frac{\cot x - \tan x}{\cos} x \sin x = {\csc}^{2} x - {\sec}^{2} x$

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

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

$1 - \left({\sin}^{2} \frac{x}{\cos} ^ 2 x\right) = {\csc}^{2} x - {\sec}^{2} x$

$1 - {\tan}^{2} x = {\csc}^{2} x - {\sec}^{2} x$

color(red)("recall the identity" tan^2x=sec^2x-1

$1 - \left({\sec}^{2} x - 1\right) = {\csc}^{2} x - {\sec}^{2} x$

$1 - {\sec}^{2} x + 1 = {\csc}^{2} x - {\sec}^{2} x$

$2 - {\sec}^{2} x \ne {\csc}^{2} x - {\sec}^{2} x$

So as written the identity doesn't work. I suspect that what was meant was for the left hand $\sin x$ term to be in the denominator. Let's try that:

$\frac{\cot x - \tan x}{\cos x \sin x} = {\csc}^{2} x - {\sec}^{2} x$

$\cot \frac{x}{\cos x \sin x} - \tan \frac{x}{\cos x \sin x} = {\csc}^{2} x - {\sec}^{2} x$

$\frac{\cos \frac{x}{\sin} x}{\cos x \sin x} - \frac{\sin \frac{x}{\cos} x}{\cos x \sin x} = {\csc}^{2} x - {\sec}^{2} x$

$\cos \frac{x}{\sin x \cos x \sin x} - \sin \frac{x}{\cos x \cos x \sin x} = {\csc}^{2} x - {\sec}^{2} x$

$\cancel{\cos} \frac{x}{\sin x \cancel{\cos} x \sin x} - \cancel{\sin} \frac{x}{\cos x \cos x \cancel{\sin} x} = {\csc}^{2} x - {\sec}^{2} x$

$\frac{1}{{\sin}^{2} x} - \frac{1}{{\cos}^{2} x} = {\csc}^{2} x - {\sec}^{2} x$

${\csc}^{2} x - {\sec}^{2} x = {\csc}^{2} x - {\sec}^{2} x$