# Question #f68d0

Mar 23, 2017

$\sec x - \sin x \tan x$

#### Explanation:

Since $\csc x = \frac{1}{\sin} x$ and $\cot x = \cos \frac{x}{\sin} x$, substitute these in to get $\frac{\frac{1}{\sin} x - \sin x}{\cos \frac{x}{\sin} x}$

Multiply the reciprocal of the denominator.

$\frac{1}{\sin} x - \sin x \cdot \sin \frac{x}{\cos} x$

We end up with this after using the distributive property.

$\sin \frac{x}{\sin x \cos x} - {\sin}^{2} \frac{x}{\cos} x$

Cancel out the two $\sin x$s on the left.

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

Since $\frac{1}{\cos} x = \sec x$ and $\sin \frac{x}{\cos} x = \tan x$, the simplified answer is

$\sec x - \sin x \tan x$

EDIT: Ignore this, not fully simplified. See Scott's answer.

Mar 23, 2017

$\cos x$

#### Explanation:

First put everything in terms of sine and cosine

$\csc x = \frac{1}{\sin} x$ and $\cot x = \cos \frac{x}{\sin} x$, so $\frac{\csc x - \sin x}{\cot} x = \frac{\frac{1}{\sin} x - \sin x}{\cos \frac{x}{\sin} x} = \left(\frac{1}{\sin} x - \sin x\right) \left(\sin \frac{x}{\cos} x\right)$

Then distribute multiplication over subtraction

$\left(\frac{1}{\sin} x - \sin x\right) \left(\sin \frac{x}{\cos} x\right) = \left(\frac{1}{\sin} x \cdot \sin \frac{x}{\cos} x\right) - \left(\sin x \cdot \sin \frac{x}{\cos} x\right)$

Multiply inside the parentheses

$\left(\frac{1}{\sin} x \cdot \sin \frac{x}{\cos} x\right) - \left(\sin x \cdot \sin \frac{x}{\cos} x\right) = \left(\sin \frac{x}{\sin x \cos x}\right) - \left(\frac{\sin x \sin x}{\cos} x\right)$

And simplify

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

Using the Pythagorean identity ${\sin}^{2} x + {\cos}^{2} x = 1$, we know that $1 - {\sin}^{2} x = {\cos}^{2} x$, so substitute that in

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

And finally, simplify

${\cos}^{2} \frac{x}{\cos} x = \cos \frac{x}{1} = \cos x$