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$$\csc(x)- \cot(x)= \frac{\sin(x)}{ 1+ \cos(x)}$$

I'm completely stumped. There are a few examples with the signs reversed but this is just different enough that none of the examples work. Is this a typo or am I just not getting it? Step-by-step would be helpful.

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Express the left side as $\frac{1}{\sin x}-\frac{\cos x}{\sin x}$, that is, as $\frac{1-\cos x}{\sin x}$. Then multiply top and bottom by $1+\cos x$, and use the identity $1-\cos^2 x=\sin^2 x$.

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Thanks. That worked out nicely. – Jim Irwin Jul 27 '14 at 0:34
You are welcome. There are other ways. However, since there were sines and cosines on the right, it seemed sensible to change the less familiar functions $\csc$ and $\cot$ to sines and cosines. – André Nicolas Jul 27 '14 at 0:37
I'll just mention that this has connections to a couple other significant identities that you'll be learning shortly, if you haven't already: $$ \tan \frac{x}{2} \ = \ \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}} \ \frac{2 \cos \frac{x}{2}}{2\cos \frac{x}{2}} \ = \ \frac{2 \ \sin \frac{x}{2} \ \cos \frac{x}{2}}{2 \ \cos^2 \frac{x}{2}} $$ $$ = \frac{\sin \ (2 \cdot \frac{x}{2})}{2 \ ( \frac{1}{2} [1 \ + \ \cos (2 \cdot \frac{x}{2}) ] )} \ = \ \frac{\sin \ x}{1 + \cos \ x} \ = \ \frac{1 - \cos \ x}{\sin \ x} \ \ . $$ So $ \ \csc x \ - \ \cot x \ = \ \tan \frac{x}{2} \ $ in all its forms. – RecklessReckoner Jul 27 '14 at 2:10
Chances are that regrettably the nice identities involving $\tan(x/2)$ are not part of the curriculum. – André Nicolas Jul 27 '14 at 2:24

$$\sin^2x=1-\cos^2x=(1+\cos x)(1-\cos x)$$

$$\iff\frac{\sin x}{1+\cos x}=\frac{1-\cos x}{\sin x}=\frac1{\sin x}-\frac{\cos x}{\sin x}=\cdots$$

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