I think I understand this but just want to confirm that I have simplified sufficiently.
Given: $$g(x)=\sin^2x$$ $$h(x)=\cot x$$$$f(x)=g(x)-h(x)$$ I am asked to find $f(x)$ and simplify:
\begin{equation} \begin{split} f(x)& = g(x)-h(x) \\ & = \sin^2x-\cot x\\ &=\dfrac{\sin^2x(\sin x)}{1(\sin x)}-\left(\dfrac{\cos x}{\sin x}\right)\\ &=\dfrac{\sin^3x-\cos x}{\sin x}\\ &=\boxed{cscx(\sin^3x-cosx)} \end{split} \end{equation}
Given: $$f(x)=\sin^2x$$$$g(x)=\cot x$$ Determine the equation $\left(\dfrac{f}{g}\right)(x)$: \begin{equation} \begin{split} \left(\dfrac{f}{g}\right)(x)&=\dfrac{\sin^2x}{\cot x}\\ &=\dfrac{\sin^2x}{\dfrac{\cos x}{\sin x}}\\ &=\dfrac{\sin^2x}{1}\cdot\dfrac{\sin x}{\cos x}\\ &=\dfrac{\sin^3}{\cos x}\\ &=\boxed{\sin^2x\tan x} \end{split} \end{equation}