Simplify

Hello, mortyr!

The first two are just "algebra": multiplying, cancelling, and combining.

Quote:

$C)\;\;\frac{\cot x}{\sec x - \tan x} - 1$

We have: . $\frac{\cot x - \sec x + \tan x}{\sec x - \tan x} \;=\;\frac{\frac{\cos x}{\sin x} - \frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x} - \frac{\sin x}{\cos x}}$

Multiply top and bottom by $\sin x\cos x\!:\;\;\frac{\sin x\cos x}{\sin x\cos x}\cdot\frac{\frac{\cos x}{\sin x} - \frac{1}{\cos x} + \frac{\sin x}{\cos x}}{\frac{1}{\cos x} - \frac{\sin x}{\cos x}}$

and we have: . $\frac{\cos^2x - \sin x + \sin^2x}{\sin x - \sin^2x}\;=\;\frac{1 - \sin x}{\sin x(1 - \sin x)}$

Reduce: . $\frac{1}{\sin x} \;=\;\boxed{\csc x}$

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Here's another approach . . .
. . guarenteed to impress/surprise/terrify your teacher.

Multiply top and bottom of the fraction by $\sec x + \tan x\!:$

. . $\frac{\cot x}{\sec x -\tan x}\cdot\frac{\sec x + \tan x}{\sec x + \tan x} - 1 \;=\;\frac{\cot x(\sec x + \tan x)}{\sec^2x - \tan^2x} - 1$

Since $\sec^2x - \tan^2x\:=\:1$ , we have: . $\cot x(\sec x + \tan x) - 1$

Then we have: . $\cot x\sec x + \cot x\tan x - 1$

Since $\cot x\tan x = 1$ , we have: . $\cot x\sec x + 1 - 1 \;=\;\cot x\sec x$

And we have: . $\frac{\cos x}{\sin x}\cdot\frac{1}{\cos x}\;=\;\frac{1}{\sin x}\;=\;\csc x$