Proving $\frac{\csc x + \cot x}{\tan x + \sin x} = \cot x\csc x$

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.

• Note that $\csc x + \cot x = \dfrac{1}{\sin x} + \dfrac{\cos x}{\sin x} = \dfrac{1 + \cos x}{\sin x}$ – Chaitanya Tappu Nov 30 '18 at 2:29

$$\require{cancel}$$ As Chaitanya Tappu noted, you made a mistake when adding $$\csc x$$ and $$\cot x$$.
$$\frac{\csc x+\cot x}{\tan x+\sin x}=\frac{\frac{1}{\sin x}+\frac{\cos }{\sin x}}{\frac{\sin x}{\cos x}+\frac{\sin x\cos x}{\cos x}}=\frac{\frac{1+\cos x}{\sin x}}{\frac{\sin x(1+\cos x)}{\cos x}}=\frac{\cancel{1+\cos x}}{\sin x}\cdot\frac{\cos x}{\sin x\cancel{(1+\cos x)}}$$ $$=\frac{\cos x}{\sin x}\cdot\frac{1}{\sin x}=\cot x\csc x$$
$$\dfrac{a+b}{\dfrac1a+\dfrac1b}=\cdots=ab$$ for $$a+b\ne0$$
$$\tan x=\dfrac1?,\sin x=\dfrac1?$$