I'm stuck at integrating an arc length over a circle + angle between curves Let $(\phi, \theta)$ be the usual spherical coordinates. And let $\gamma (t) = (\phi (t), \theta (t))$ be the curve given by
$$ \phi (t) = \ln\left(\cot\left(\frac{\pi}{4} - \frac{t}{2}\right)\right),$$ $$ \theta (t) = \frac{\pi}{2} - t.$$
$t \in (0, \frac{\pi}{2})$

a) Compute the arc length of $ \gamma (t). $
b) Show that the curve  cuts each "parallel" in a constant angle, by computing $(\gamma (t))'$ and $\psi_{\phi}$
c) Make a diagram of the curves

So the formula for arc length would be $$ \begin{equation}
\ell_\gamma = \int_0^1 \sqrt{ \left ( \frac{dr}{dt} \right )^2 +
  r^2\sin^2 \theta \left ( \frac{d \phi}{dt} \right)^2  + 
    r^2 \left ( \frac{d \theta}{dt} \right )^2} \, dt.
\end{equation}$$
In my case, as  $$\phi'(t) = \frac 1 {2\cos\left(\frac{\pi}{4} - \frac{t}{2}\right) \sin\left(\frac{\pi}{4} - \frac{t}{2}\right)}, $$  and $$ \theta ' (t) = -1 ,$$ then 
\begin{align*}
\ell_\gamma &= \int_0^1 \sqrt{
  \sin^2 \theta \left ( \frac{d \phi}{dt} \right)^2  + 
    \left ( \frac{d \theta}{dt} \right )^2} \, dt \\
&= \int_0^1 \sqrt{ \sin^2 \theta \left (\frac 1 {2\cos(\frac{\pi}{4} - \frac{t}{2})\sin(\frac{\pi}{4} - \frac{t}{2})} \right)^2  + 1} \, dt\\
&= \int_0^1 \sqrt{
  \left(\frac{\sin \theta }{2\cos(\frac{\pi}{4} - \frac{t}{2})\sin(\frac{\pi}{4} - \frac{t}{2})} \right)^2  + 
      1} \, dt.
\end{align*}
I am stuck here. Before I go further, would this integral be correct? That $\sin^2(\theta)$ confuses me maybe, I don't know if I should change the argument to something else.
For exercise b) Could someone verify this reasoning?
I want the angle between two curves on the sphere, which is defined as the angle between the tangent vectors of this curves at the point of intersection. So I want to find $\alpha$ such that $\alpha = cos^{-1}(\frac{<\beta '(t), \gamma ' (t)>}{||\beta '(t) || ||\gamma ' (t)||}) $, because that is what the angle between two vectors in a vector space is.
So if $\beta (t) = (t, \theta_0)$ is a curve that describes the parallel curve in a sphere, with $\theta_0$ constant, Its tangents are given by $(1, 0),$. I already have the derivatives of $\gamma$ so $$ \alpha = cos^{-1}(\frac{<\beta '(t), \gamma ' (t)>}{||\beta '(t) || ||\gamma ' (t)||}) = cos^{-1}(\frac{<(1, 0) \dot (\frac{1}{sin{\frac{\pi}{2}- t}}, -1)>}{(1) \sqrt{csc^2(\frac{\pi}{2}-t) + 1}})$$
$$ = cos^{-1}(\frac{1}{sin{\frac{\pi}{2}- t}}{ \sqrt{csc^2(\frac{\pi}{2}-t) + 1}})$$.
Thanks in advance.
 A: \begin{align}
& \int_0^1 \sqrt{ \left(\frac{\sin (\frac{\pi}{2} - t) }{2\cos(\frac{\pi}{4} - \frac t 2) \sin(\frac \pi 4 - \frac t 2)} \right)^2 + 1} \, dt \\[10pt]
\end{align}
As $$\sin^2(\frac \pi 4 - \frac t 2)cos^2(\frac \pi 4 - \frac t 2) = \frac{1 - \cos(\pi - 2t)}{8},$$ then
\begin{align}
& \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{1 -\cos(2(\frac{\pi}{2} -  t ))}\right) + 1} \, dt & = \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{1 -[\cos^2(\frac{\pi}{2} -  t )- \sin^2(\frac{\pi}{2} -  t ))]}\right) + 1} \, dt  \end{align}
\begin{align} = \int_0^1 \sqrt{ \left(\frac{2\sin^2 (\frac{\pi}{2} - t) }{2\sin^2(\frac{\pi}{2} - t)}\right) + 1} \, dt  = \sqrt2 \\[10pt] 
\end{align}
A: \begin{align}
& \int_0^1 \sqrt{ \left(\frac{\sin \theta }{2\cos(\frac{\pi}{4} - \frac t 2) \sin(\frac \pi 4 - \frac t 2)} \right)^2 + 1} \, dt \\[10pt]
= {} & \int_0^1 \sqrt{\left(\frac{\sin\theta}{\cos t} \right)^2+1} \, dt = \int_0^1 \sqrt{ \left( \frac{\cos t}{\cos t} \right)^2 + 1} \, dt = \int_0^1 \sqrt 2 \, dt = \sqrt 2.
\end{align}
The first two equalities are trigonometric identities:
\begin{align}
& 2\cos\left( \frac \pi 4 - \frac t 2 \right) \sin \left( \frac \pi 4 - \frac t 2 \right) = \sin\left( \frac \pi 2 - t \right) = \cos t \\[10pt]
& 
\end{align}
