Trigonometry (value of expression) I have been revising trigonometry and I couldn't solve this particular one from my book.
One has to find the value of given expression 
$\frac{tan\alpha}{1-cos\alpha}$, if $sin\alpha=-\frac{2}{3}$ and $ctn\alpha&gt;0$.

I tried expressing every trigonometric function through $sin\alpha$ by multiplying numerator and denominator by $cos\alpha$. That would yield $sin\alpha$ in numerator and $cos\alpha$ functions in denominator.

Then I tried expanding further:

Since it was given that $ctan\alpha$>0, then $ctn\alpha=\frac{cos\alpha}{sin\alpha} \Rightarrow cos\alpha=ctn\alpha \times sin\alpha \Rightarrow \pm \sqrt{\frac{1}{sin^2\alpha}-1} \times sin\alpha= \sqrt{\frac{1}{sin^2\alpha}-1} \times sin\alpha$

I suppose that the value of $ctn\alpha$ will have to be positive.

$\frac{sin\alpha}{cos\alpha+cos^2\alpha}=\frac{sin\alpha}{\sqrt{\frac{1}{sin^2\alpha}-1} \times sin\alpha-1+sin^2\alpha}$

When I input initial condition for sine value, I still somehow can't get the result stated in my book (answers are given in random order; my answer doesn't quite match any of them)

I would appreciate any help with this!
HINT:

As $\dfrac{\cos\alpha}{\sin\alpha}=\cot\alpha&gt;0,\cos\alpha&lt;0\implies\cos\alpha=-\sqrt{1-\sin^2\alpha}$

$\tan\alpha=\dfrac{\sin\alpha}{\cos\alpha}=?$
Thanks, both respondents!
I don't know why I've overcomplicated this so much for myself.
With both methods, I finally managed to get the result of $\frac {3(3 \sqrt {5}+5)}{10}$, which matches the given set of answers.