1. ## Prove/show question

Heya!

I've been struggling with this for some time now and would really appreciate some help. Need to show that the right side equals the left side.

$\displaystyle \frac{1}{sinx} = \frac{sinx}{1+cosx} + \frac{1}{tanx}$

Tried different ways to solve this but I get stuck each time.

RS = Rightside; sorry if this is not the right way to type this. Not familiar with the english terms

RS: $\displaystyle \frac{sinx}{1+cosx} + \frac{1}{\frac{sinx}{cosx}}$

After this I've experimented with changing 1's to $\displaystyle sin^2x+cos^2x$ among many other things.

Honestly feels like I am really on the wrong track here. Mostly the $\displaystyle 1+cosx$ that annoys me.

Can someone shed some light on this for me? Thanks in advance!

2. $\displaystyle \frac{\sin{x}}{1+\cos{x}} + \frac{\cos{x}}{\sin{x}}$

$\displaystyle \frac{\sin{x}}{1+\cos{x}} \cdot \frac{1-\cos{x}}{1-\cos{x}} + \frac{\cos{x}}{\sin{x}}$

$\displaystyle \frac{\sin{x}(1-\cos{x})}{1-\cos^2{x}} + \frac{\cos{x}}{\sin{x}}$

$\displaystyle \frac{\sin{x}(1-\cos{x})}{\sin^2{x}} + \frac{\cos{x}}{\sin{x}}$

$\displaystyle \frac{1-\cos{x}}{\sin{x}} + \frac{\cos{x}}{\sin{x}}$

$\displaystyle \frac{1 - \cos{x} + \cos{x}}{\sin{x}}$

$\displaystyle \frac{1}{\sin{x}}$

3. Thank you skeeter.

Didn't realize 1/tan = cotx = cosx/sinx. When knowing this it was pretty simple. Cheers!