# How do you prove csc (-x) / sec ( - x) =- cot x?

Mar 17, 2018

We're trying to prove that $\csc \frac{- x}{\sec} \left(- x\right) = - \cot x$.

You'll need to use reciprocal identities:

$\sec x = \frac{1}{\cos} x$

$\csc x = \frac{1}{\sin} x$

and also angle difference formulae:

$\sin \left(x - y\right) = \sin x \cos y - \cos x \sin y$

$\cos \left(x - y\right) = \cos x \cos y + \sin x \sin y$

Here's the actual problem. I'll be manipulating the right side of the equation until it equals the right:

$L H S = \csc \frac{- x}{\sec} \left(- x\right)$

color(white)(LHS)=(quad1/sin(-x)quad)/(1/(cos(-x))

$\textcolor{w h i t e}{L H S} = \frac{1}{\sin} \left(- x\right) \cdot \cos \frac{- x}{1}$

$\textcolor{w h i t e}{L H S} = \cos \frac{- x}{\sin} \left(- x\right)$

$\textcolor{w h i t e}{L H S} = \cos \frac{0 - x}{\sin} \left(0 - x\right)$

$\textcolor{w h i t e}{L H S} = \frac{\cos 0 \cos x + \sin 0 \sin x}{\sin 0 \cos x - \cos 0 \sin x}$

$\textcolor{w h i t e}{L H S} = \frac{1 \cdot \cos x + 0 \cdot \sin x}{0 \cdot \cos x - 1 \cdot \sin x}$

$\textcolor{w h i t e}{L H S} = \frac{1 \cdot \cos x + \textcolor{red}{\cancel{\textcolor{b l a c k}{0 \cdot \sin x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{0 \cdot \cos x}}} - 1 \cdot \sin x}$

$\textcolor{w h i t e}{L H S} = \frac{1 \cdot \cos x}{- 1 \cdot \sin x}$

$\textcolor{w h i t e}{L H S} = \cos \frac{x}{- 1 \cdot \sin x}$

$\textcolor{w h i t e}{L H S} = - 1 \cdot \cos \frac{x}{\sin} x$

$\textcolor{w h i t e}{L H S} = - 1 \cdot \cot x$

$\textcolor{w h i t e}{L H S} = - \cot x$

$\textcolor{w h i t e}{L H S} = R H S$

That's the proof. Hope this helped!