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Prove $$ \frac{\sin\theta}{1-\cos\theta} - \frac{\sin\theta}{1+\cos\theta} = 2\cot \theta$$

So I started by combining the two fractions, which gave me: $$ \frac{\sin\theta(1+\cos\theta) - \sin\theta(1-\cos\theta)}{(1-\cos\theta)(1+\cos\theta)} = \frac{2\sin\theta\cos\theta}{1-\cos^2\theta} = \frac{\sin2\theta}{1-\cos^2\theta}$$ I wasn't sure where to go from here considering I'm aiming for $2\cos\theta / \sin\theta$

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  • $\begingroup$ Just use $\sin(2\theta) = 2 \sin(\theta) \cos(\theta)$ and $1-\cos^2(\theta) = \sin^2(\theta)$. $\endgroup$ – TheSilverDoe Apr 14 at 19:48
  • $\begingroup$ Why are you writing $\sin2\theta$ instead of simply changing $1-\cos^2\theta=\sin^2\theta$ and simplify? $\endgroup$ – egreg Apr 14 at 20:26
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$$\frac{2\sin\theta\cos\theta}{1-\cos^2\theta} =\frac{2\sin\theta\cos\theta}{\sin^2\theta} = \frac{2\cos\theta}{\sin \theta} = 2\cot \theta$$

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    $\begingroup$ @H.Linkhorn, Maria Mazur has provided the answer. I wonder why some people down voted it! I up voted! Note that the LHS in Maria's answer is the simplification (just cross multiplication) of the LHS in your original question. $\endgroup$ – ersh Apr 14 at 20:01
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It's $$\frac{\sin2\theta}{\sin^2\theta}=\frac{2\sin\theta\cos\theta}{\sin^2\theta}=\frac{2\cos\theta}{\sin\theta}=2\cot\theta.$$

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  • $\begingroup$ Where does that identity come from? $\endgroup$ – H.Linkhorn Apr 14 at 19:47
  • $\begingroup$ @H.Linkhorn I added something. See now. $\endgroup$ – Michael Rozenberg Apr 14 at 19:49
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As $\sin^2t=1-\cos^2t=(1-?)(1+?)$

$$\dfrac{\sin t}{1\pm\cos t}=\dfrac{1\mp \cos t}{\sin t}$$

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