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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.

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    $\begingroup$ Note that $\csc x + \cot x = \dfrac{1}{\sin x} + \dfrac{\cos x}{\sin x} = \dfrac{1 + \cos x}{\sin x}$ $\endgroup$ – Chaitanya Tappu Nov 30 '18 at 2:29
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$\require{cancel}$ As Chaitanya Tappu noted, you made a mistake when adding $\csc x$ and $\cot x$.

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

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    $\begingroup$ you beat me to it. (+1) $\endgroup$ – clathratus Nov 30 '18 at 2:37
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$$\dfrac{a+b}{\dfrac1a+\dfrac1b}=\cdots=ab$$ for $a+b\ne0$

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

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