# How do you prove (sinx + cosxcotx)/cotx = secx?

May 17, 2015

We need to get the left side equal to the right side:

So we will look at $= \frac{\sin x + \cos x \cot x}{\cot} x$

we can use the identity of $\textcolor{red}{\cot x = \cos \frac{x}{\sin} x}$ here

so we will have $= \frac{\sin x + \cos x \left(\cos \frac{x}{\sin} x\right)}{\cos \frac{x}{\sin} x}$

if we now take out a $\textcolor{red}{\frac{1}{\sin} x}$ from the top, we will have:

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

there we can use the identity that $\textcolor{red}{{\sin}^{2} x + {\cos}^{2} x = 1}$

thus we are left with $= \frac{\frac{1}{\sin} x}{\cos \frac{x}{\sin} x}$

which we can write as $= \left(\frac{1}{\sin} x\right) \left(\sin \frac{x}{\cos} x\right)$

the $\textcolor{red}{\sin x}$'s will cancel each other, and we will be left with:

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

which is our identity to $\textcolor{red}{\left(\frac{1}{\cos} x\right) = \sec x}$

thus we get $= \sec x$

and we have that the right hand side $=$ to left hand side