<s>[INST] Find the first derivative $y=\sqrt\frac{1+\cosθ}{1-\cosθ}$ $$y=\sqrt\frac{1+\cosθ}{1-\cosθ}$$ my professor said that the answer is $$y'=\frac{1}{\cosθ-1}$$ she said use half angle formula but I just end up with $\frac{(-2\sinθ)\sqrt{(1-\cosθ)(1+\cosθ)}}{2(1-\cosθ)^2(1+\cosθ)}$ I used the quotient rule. I also know that ${(1-\cosθ)^2}$ can be $\sinθ^2$ but i try to apply my identities but it's still wrong.
 [/INST] $$ y^2=\frac {1+\cos\theta}{1-\cos\theta}.$$
Differentiating gives
$$\frac {d}{d\theta} y^2=\frac {d}{d\theta} \frac {1+\cos\theta}{1-\cos\theta} = -\frac{2\sin\theta}{(1-\cos\theta)^2}. $$
Using the chain rule gives for the left hand side
$$2y\frac {dy}{d\theta}. $$ Hence, $$\frac {dy}{d\theta}=-\frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{(1-\cos\theta)^2},$$ a plot of which agrees with that of Jean-Claude Arbaut's answer.
Assuming this to be the case then retrospectively it would seem that
$$\text{sgn}(\sin\theta) = \frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{1-\cos\theta}.$$
Interesting.

Out of curiosity, let $\theta= \arcsin x$. Then
$$\frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{1-\cos\theta}=\frac{x}{\sqrt{1-\sqrt{1-x^2}} \sqrt{\sqrt{1-x^2}+1}}=\frac{x}{\sqrt{x^2}}=\frac{x}{\mid x\mid}=\text{sgn}(x) .$$
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