2. No, but you can multiply the top and bottom of $\displaystyle \frac{\cos{x}}{1 - \sin{x}}$ by $\displaystyle 1 + \sin{x}$...
3. $\dfrac{1+\sin(x)}{\cos(x)} = \dfrac{1+\sin(x)}{\cos(x)} \times \dfrac{1-\sin(x)}{1-\sin(x)} =\dfrac{1-\sin^2(x)}{\cos(x)(1-\sin(x))}= ...$