$\frac{\cos\theta }{1-\sin\theta }\cdot \frac{1+\sin\theta }{1+\sin\theta }$

Question

I'm having trouble can someone help and explain step by step on how I can simplify the answer?

Answer 1

HINT

You have $a^2-b^2 = (a-b)(a+b)$ and $\sin^2 x + \cos^2 x = 1$, therefore $$ \begin{split} \frac{\cos t}{1-\sin t}\cdot \frac{1+\sin t}{1+\sin t} &= \frac{\cos t(1+\sin t)}{(1-\sin t)(1+\sin t)} \\ &= \frac{\cos t(1+\sin t)}{1^2-\sin^2 t} \\ &= \frac{\cos t(1+\sin t)}{\cos^2 t} \\ \end{split} $$

Can you finish?