# How do you prove (secx)/(sinx)-(sinx)/(cosx)=cotx?

May 3, 2018

$L H S = \frac{\sec x}{\sin x} - \frac{\sin x}{\cos x}$

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

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

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

$= \cos \frac{x}{\sin} x$

$= \cot x = R H S$

May 4, 2018

$\sec \frac{x}{\sin} x - \sin \frac{x}{\cos} x = \cot x$

I'll be simplifying down the LHS of the equation to get to the RHS of the equation. The $\textcolor{b l u e}{\text{blue color}}$ refers to what is being changed.

First, we know that $\textcolor{b l u e}{\sin \frac{x}{\cos} x}$ is the same as $\textcolor{b l u e}{\tan x}$:
$\sec \frac{x}{\sin} x - \textcolor{b l u e}{\tan x}$

Now, multiply $\textcolor{b l u e}{\sin \frac{x}{\sin} x}$ to $\textcolor{b l u e}{\tan x}$ so that both expressions have the same denominator:
$\sec \frac{x}{\sin} x - \frac{\tan x \textcolor{b l u e}{\sin x}}{\textcolor{b l u e}{\sin x}}$

Combine both expressions to one denominator:
$\frac{\sec x - \tan x \sin x}{\sin} x$

We know that $\textcolor{b l u e}{\sec x = \frac{1}{\cos} x}$ and $\textcolor{b l u e}{\tan x = \sin \frac{x}{\cos} x}$:
$\frac{\textcolor{b l u e}{\frac{1}{\cos} x} - \textcolor{b l u e}{\sin \frac{x}{\cos} x} \cdot \sin x}{\sin} x$

Multiply $\textcolor{b l u e}{\sin} x$ with $\textcolor{b l u e}{\sin} x$:
$\frac{\frac{1}{\cos} x - \frac{\textcolor{b l u e}{{\sin}^{2} x}}{\cos} x}{\sin} x$

Combine numerator and denominator (from the top)
$\frac{\frac{\textcolor{b l u e}{1 - {\sin}^{2} x}}{\cos} x}{\sin} x$

From the Pythagorean Identities, we know that $\textcolor{b l u e}{1 - {\sin}^{2} x = {\cos}^{2} x}$:
$\frac{\textcolor{b l u e}{{\cos}^{2} \frac{x}{\cos} x}}{\sin} x$

Divide $\textcolor{b l u e}{{\cos}^{2} \frac{x}{\cos} x}$:
$\frac{\textcolor{b l u e}{\cos \frac{x}{1}}}{\sin} x$

Simplify:
$\frac{\textcolor{b l u e}{\cos x}}{\sin} x$

We know that $\textcolor{b l u e}{\cos \frac{x}{\sin} x = \cot x}$:
$\textcolor{b l u e}{\cot} x$

Now, the left hand side is equivalent to the right hand side, so we have proven that $\sec \frac{x}{\sin} x - \sin \frac{x}{\cos} x = \cot x$.

Hope this helps!