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I want to simplify this expression to contain 2 or fewer terms. Is this possible, and if so, what are the steps I can take to do so?

sin x / ( 1 + cos x )

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    You may need to use identities at best. Try conjugation: multiplying and dividing by $1 - \cos x$. – 2017-01-31
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    A sketch makes it possible in this case to guess the answer – 2017-01-31
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    The expression is already quite simple. What do you call "simplify" ? – 2017-01-31
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    @YvesDaoust To simplify an expression is to convert it into less terms. Thus, 1 or 2 terms would be a simplified version of this expression :) – 2017-01-31
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    @GiantSpruce: would you consider $\csc x-\cot x$ to be simpler ? – 2017-01-31
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    @YvesDaoust Yes – 2017-01-31
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    @YvesDaoust I have edited my post to better clarify the question. Please provide me with feedback so I can improve it and ask better ones in the future. – 2017-01-31
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    @GiantSpruce: technically speaking, the original expression has only one *term*, which is the quotient of two *factors*. – 2017-01-31
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52827/discussion-between-giantspruce-and-yves-daoust). – 2017-01-31

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I have another idea $1+\cos x=2\cos^2 {\frac{x}{2}}$ and $\sin x=2\sin{\frac{x}{2}}\cos{\frac{x}{2}}$.

Now, the given can be written as $\tan {\frac{x}{2}}$.

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    :D that's also very nice! – 2017-01-31
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$$\frac{\sin(x)}{1+\cos(x)}=\frac{\sin(x)}{1+\cos(x)}\frac{1-\cos(x)}{1-\cos(x)}=\frac{\sin(x)(1-\cos(x))}{1-\cos^2(x)}$$

Recall the pythagorean identity to see that $1-\cos^2(x)=\sin^2(x)$.

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

$$=\frac1{\sin(x)}-\frac{\cos(x)}{\sin(x)}$$

$$=\csc(x)-\cot(x)$$

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Hint The appearance of $\color{red}{1 + \cos x}$ suggests we can produce an expression without a constant term in the denominator by substituting $x = 2t$ and using the half-angle identity $\cos^2 t = \frac{1}{2}(\color{red}{1 + \cos 2 t})$.

Substituting in our expression gives $$\frac{\sin x}{1 + \cos x} = \frac{\sin 2t}{1 + \cos 2 t} = \frac{2 \sin t \cos t}{2 \cos^2 t} = \frac{\sin t}{\cos t} = \tan t = \tan \frac{x}{2} .$$

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    @Henry Thanks! I've corrected the answer---thanks to the Pythagorean identity, the same technique works just as well for the correct sign, too. – 2017-01-31
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Use the formulae:

sin x= (2 tan x/2)/ (1+(tanx/2)^2)

cos x= ( 1-(tan x/2)^2)/ (1+(tan x/2)^2)