Is $\frac{(|x| - x^2)}{\sin x}$ an odd function, if it is odd, how do I prove that?

Question

This is the function:

$$f(x)=\frac{|x|-x^{2}}{\sin x}$$

I am not understanding where to go from the second line:

$$\begin{array}{l} f(-x) & =\displaystyle\frac{|-x|-(-x)^{2}}{\sin(-x)}\\ \displaystyle & =\displaystyle\frac{|-x|-(-x)^{2}}{-\sin x} \end{array}$$

How do make this equal $-f(x)$? Or is it neither odd nor even?

Answer 1

We have $|-x|=|x|$ and $(-x)^2=x^2$.

Answer 2

You have $$\begin{array}{l} f(- x) =\frac{|x|\ -\ ( x)^{2}}{-sin( x)}\\ \end{array}$$ and this is $-f(x)$ so $f$ is odd.