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

Question

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.

Answer 1

$\require{cancel}$ As Chaitanya Tappu noted, you made a mistake when adding $\csc x$ and $\cot x$.

$$\frac{\csc x+\cot x}{\tan x+\sin x}=\frac{\frac{1}{\sin x}+\frac{\cos }{\sin x}}{\frac{\sin x}{\cos x}+\frac{\sin x\cos x}{\cos x}}=\frac{\frac{1+\cos x}{\sin x}}{\frac{\sin x(1+\cos x)}{\cos x}}=\frac{\cancel{1+\cos x}}{\sin x}\cdot\frac{\cos x}{\sin x\cancel{(1+\cos x)}}$$ $$=\frac{\cos x}{\sin x}\cdot\frac{1}{\sin x}=\cot x\csc x$$

Answer 2

$$\dfrac{a+b}{\dfrac1a+\dfrac1b}=\cdots=ab$$ for $a+b\ne0$

$\tan x=\dfrac1?,\sin x=\dfrac1?$