I'm a little confused how to start this problem:
1 + cosx + sinx
sinx 1 + cosx
Lauren you have to be a little more clear on your problem. Are we proving this identity to be something else? It just looks like a whole lot of cos's and sin's thrown onto the screen at the moment.
Could it be like this?
$\displaystyle \frac{1+cos(x)+sin(x)}{sin(1+cos(x))} = \cdots$
Looks like a tricky one, these proofs can be unforgiving at times but if you get one to fall out it feels great.
Have a look at this list of identities
Table of Trigonometric Identities
Then remember that
$\displaystyle \frac{1}{sin(x)} = cosec(x)$
Try try to put everything one the left hand side in terms of sin(x)
please use grouping symbols to make your post clear ...
(1+cosx)/sinx + sinx/(1+cosx)
$\displaystyle \frac{1+\cos{x}}{\sin{x}} + \frac{\sin{x}}{1+\cos{x}} = 2\csc{x}$
$\displaystyle \frac{1+\cos{x}}{\sin{x}} + \frac{\sin{x}}{1+\cos{x}} \cdot \frac{1-\cos{x}}{1-\cos{x}} =$
$\displaystyle \frac{1+\cos{x}}{\sin{x}} + \frac{\sin{x}(1-\cos{x})}{1-\cos^2{x}} =
$
$\displaystyle \frac{1+\cos{x}}{\sin{x}} + \frac{\sin{x}(1-\cos{x})}{\sin^2{x}} =$
$\displaystyle \frac{1+\cos{x}}{\sin{x}} + \frac{1-\cos{x}}{\sin{x}} =$
$\displaystyle \frac{1+\cos{x} + 1 - \cos{x}}{\sin{x}} =$
$\displaystyle \frac{2}{\sin{x}} = 2\csc{x}$