I've tried many things, can't seem to work it out. Thank you!
$\displaystyle \frac{{\tan \theta - \sin \theta }}
{{\sin ^3 \theta }} = \frac{{\sin \theta \left( {\dfrac{1}
{{\cos \theta }} - 1} \right)}}
{{\sin ^3 \theta }} = \frac{{1 - \cos \theta }}
{{\cos \theta (1 + \cos \theta )(1 - \cos \theta )}}.$
Does that make sense?
$\displaystyle \frac{tan\theta - sin\theta}{sin^{3}\theta}$
1. because $\displaystyle tan\theta = \frac{sin\theta}{cos\theta}$, substitute that into $\displaystyle tan\theta$
$\displaystyle \frac{\frac{sin\theta}{cos\theta} - sin\theta}{sin^{3}\theta}$
2. Common denominator of the fractions in the numerator
$\displaystyle \frac{\frac{sin\theta}{cos\theta} - \frac{sin\theta cos\theta}{cos\theta}}{sin^{3}\theta}$
3. combine fractions in the numerator
$\displaystyle \frac{\frac{sin\theta-sin\theta cos\theta}{cos\theta}}{sin^{3}\theta}$
4. simplify (dividing a fraction by something means multiplying by it's inverse):
$\displaystyle \frac{sin\theta-sin\theta cos\theta}{cos\theta sin^{3}\theta}$
5. factor out $\displaystyle sin\theta$ in the numerator
$\displaystyle \frac{sin\theta (1 - cos\theta)}{cos\theta sin^{3}\theta}$
$\displaystyle sin\theta$ in the numerator and denominator cancel eachother out
$\displaystyle \frac{1 - cos\theta}{cos\theta sin^{2}\theta}$
6. Multiply the numerator and the denominator by $\displaystyle 1+cos\theta$
$\displaystyle \frac{(1 - cos\theta)(1 + cos\theta)}{cos\theta sin^{2}\theta (1 + cos\theta)}$
7. FOIL the numerator:
$\displaystyle \frac{1 + cos^{2}\theta}{(1 + cos\theta)cos\theta sin^{2}\theta}$
8. Using the formula $\displaystyle sin^{2}\theta + cos^{2}\theta = 1$ see that $\displaystyle sin^{2}\theta = 1 - cos^{2}\theta$ and substitute that into the equation
$\displaystyle \frac{1 - cos^{2}\theta}{(1 + cos\theta)cos\theta (1- cos^{2}\theta)}$
9. $\displaystyle 1- cos^{2}\theta$ is in the numerator and the denominator, so cancel them out
$\displaystyle \frac{1}{(1 + cos\theta)cos\theta}$