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.
 A: $$y=\sqrt{\frac{1+\cos\theta}{1-\cos\theta}}=\sqrt{\frac{2\cos^2\theta/2}{2\sin^2\theta/2}}=|\cot\theta/2|$$
Also,
$$\frac{\mathrm{d}}{\mathrm{d}\theta}\cot(\theta/2)=-\frac12\frac1{\sin^2\theta/2}$$
Hence, when $\cos\theta/2\neq0$ and $\sin\theta/2\neq 0$, that is when $\theta\notin\pi\Bbb Z$,
$$\frac{\mathrm{d}y}{\mathrm{d}\theta}=-\frac12\frac{\mathrm{sign}(\cot \theta/2)}{\sin^2\theta/2}$$
And since $2\sin^2\theta/2=1-\cos\theta$,
$$\frac{\mathrm{d}y}{\mathrm{d}\theta}=-\frac{\mathrm{sign}(\cot \theta/2)}{1-\cos\theta}$$
And yet one more simplification, when $\sin\theta/2\neq 0$,
$$\mathrm{sign}(\cot \theta/2)=\mathrm{sign}(\cos (\theta/2)\sin (\theta/2))=\mathrm{sign}(\sin\theta)$$
So, still for $\theta\notin\pi\Bbb Z$
$$\frac{\mathrm{d}y}{\mathrm{d}\theta}=-\frac{\mathrm{sign}(\sin\theta)}{1-\cos\theta}$$
Since it may not be absolutely clear, here is a plot of $y$, to show how $y$ differs from $\cot\theta/2$. Notice the angle when the curve reaches $0$: $y$ is thus not differentiable here.

A: We have 
$\dfrac{(-2\sinθ)\sqrt{(1-\cosθ)(1+\cosθ)}}{2(1-\cosθ)^2(1+\cosθ)}=-\dfrac{2\sin\theta|\sin\theta|}{2(1-\cos\theta)(1-\cos^2\theta)}$
=-sign$(\sin\theta)\dfrac1{1-\cos\theta}$
A: The half angle formulas are often useful:
$$\cos^2(x)=\dfrac{1+\cos(2x)}{2} \text{ and } \sin^2(x)=\dfrac{1-\cos(2x)}{2}$$
$$y=\sqrt{\dfrac{(1+\cos\theta)/2}{(1-\cos\theta)/2}}=\sqrt{\dfrac{\cos^2(\theta/2)}{\sin^2(\theta/2)}}=\sqrt{\cot^2(\theta/2)}=\left|\cot(\theta/2)\right|$$
The derivative of $\cot(x)$ is $(\cot(x))'=-(1+\cot^2(x))=\dfrac{-1}{\sin^2x}$
A: $\dfrac{1+\cos\theta}{1-\cos\theta}=\dfrac{(1+\cos\theta)(1-\cos\theta)}{(1-\cos\theta)^2}=\left(\dfrac{\sin\theta}{1-\cos\theta}\right)^2$
So,$\sqrt{\dfrac{1+\cos\theta}{1-\cos\theta}}=$sign of $(\sin\theta)\cdot\dfrac{\sin\theta}{1-\cos\theta}$
Now apply quotient Rule 
A: $$ y^2=\frac {1+\cos\theta}{1-\cos\theta}.$$
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
$$\frac{\sqrt{1-\cos\theta}}{\sqrt{1+\cos\theta}}\frac{\sin\theta}{1-\cos\theta}=\frac{x}{\sqrt{1-\sqrt{1-x^2}} \sqrt{\sqrt{1-x^2}+1}}=\frac{x}{\sqrt{x^2}}=\frac{x}{\mid x\mid}=\text{sgn}(x) .$$
A: Using half angle formula $$\cos\theta =2\cos^2(\theta/2)-1$$ and $$\cos\theta =1-2\sin^2(\theta/2).$$
So,
\begin{equation*}
\begin{split}
y & = \sqrt{\frac{1+\cos\theta}{1-\cos\theta}}\\
& = \sqrt{\frac{2\cos^2(\theta/2)}{2\sin^2(\theta/2)}}\\
& = \frac{\cos(\theta/2)}{\sin(\theta/2)} \quad \text{ for } 0^{\circ}\le \frac{\theta}{2}\le 90^\circ
\end{split}
\end{equation*}
Now, by using quotient rule $y'=\frac{-1/2}{\sin^2(\theta /2)}$, using half angle formula again we have $$y'=\frac{-1}{1-\cos\theta}.$$

Your professor's answer is correct provided that the denominator is $1-\cos\theta$ not $\cos\theta -1.$  
