TITLE: Proving $ \frac{\csc x + \cot x}{\tan x + \sin x} = \cot x\csc x $
QUESTION [5 upvotes]: I am currently working on understanding trig identities.
A question has me stumped, and no matter how I look at it, it never leads to the proof. I believe I am making a mistake when dividing multiple fractions.
$$ \frac{\csc x + \cot x}{\tan x + \sin x} = \cot x\csc x $$
For my first step I break up the $\csc x$ and $\cot x$ in the numerator and add them together to make:
$$\frac{\frac{1+\cos x}{\sin x\cos x}}{\tan x+\sin x}$$
I then simplify further and end up at:
$$ \frac{\cos x+\cos^2 x}{\sin^2 x\cos^2 x} $$
From here on I don't see any identities, or possible ways to decompose this further.

REPLY [3 votes]: $$\dfrac{a+b}{\dfrac1a+\dfrac1b}=\cdots=ab$$ for $a+b\ne0$
$\tan x=\dfrac1?,\sin x=\dfrac1?$