Why is this derivative incorrect? We have to find the derivative of $$f(x) = \dfrac{\tan(2x)}{\sin(x)}$$
I would like to know why my approach is incorrect:
$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)}  - \tan(2x) \cdot \cos(x)}{\sin^2(x)}$$
$$ =  \dfrac{ 2 \sin(x) - \tan(2x) \cdot \cos(x)}{\cos^2(2x) \cdot \sin^2(x)}$$
$$ = \dfrac {2 \sin(x) - \sin(2x) \cdot \cos(x)}{\cos^3(2x) \cdot \sin^2(x)}$$
p.s. - To avoid confusion ; I wanted to get rid of the $\tan$. I'm sure there is a shorter method than this but I don't want it; I just want to know why this is wrong.

$$f'(x) = \dfrac{\sin(x) \cdot \dfrac{2}{\cos^2(2x)}  - \tan(2x) \cdot \cos(x)}{\sin^2(x)}\tag{1}$$
So far so good...
But your following line is where you made an algebraic error: you forgot to multiply both (the entire numerator (both terms)) and (the denominator) by $\cos^2(2x)$. Doing this gives us: 
$$f'(x) = \dfrac{\color{blue}{\bf \cos^2(2x)}\left(\sin(x) \cdot \dfrac{2}{\cos^2(2x)}  - \tan(2x) \cdot \cos(x)\right)}{\color{blue}{\bf \cos^2(2x)}\sin^2(x)}$$
NOW we distribute across the numerator, canceling the denominator of the first term, and multiplying the second term by $\cos^2{2x}$:
$$ =  \dfrac{2\sin(x) - \tan(2x) \cdot \cos(x)\cdot \cos^2(2x)}{\cos^2(2x)\sin^2(x)}\tag{2}$$
$$ = \dfrac{2\sin(x) - \sin(2x) \cdot \cos^2(2x)\cdot \cos(x)}{\cos^3(2x)\sin^2(x)}\tag{3}$$
