Relativistic difference of angle between two inertial frame I was studying Prof. Padmanabhan's book on general relativity; "Gravitation: Foundations and Frontiers, which I found very well written. I just started the book and the exercises are also challenging.
I came across this equation in section 1.3.2, where Padmanabhan derives the direction of motion of a moving particle in different inertial frames.
The formula (eq 1.22) reads
$$
\tan \theta=\gamma^{-1}\frac{\sin \theta'}{\beta+\cos \theta'},
$$
and then he asked to show that this can be written as 
$$
\tan \frac{\theta'}{2}=e^{-\xi}\tan \frac{\theta}{2},
$$
where $\xi$ is the rapidity defined as 
$$
\tanh \xi=\beta,\quad
\gamma=\cosh \xi, \quad
\beta\gamma=\sinh\xi, \quad
\beta=\frac{V}{c},
$$
where $V$ is the magnitude of relative velocity between the two inertial frames.
It looks like a very clever manipulation of trigonometric identities which I can't figure.
Basically, I can't come up  with  $\tan \frac{\theta}{2}$ or $\tan \frac{\theta'}{2}$.
From the definition of $\xi$ some simple algebra gives you
$$
e^{-\xi}=\frac{(1-\beta)\sin\theta'}{\tan \theta(\beta+\cos \theta')},
$$
but I cant go any further. Help please.
 A: In deriving this, I found that you, or the book, have $\theta$ and $\theta'$ reversed in your second formula.
We start with the given formula for $\tan\theta$ in terms of $\theta'$, namely
$$\tan\theta=\frac{\sin\theta'}{\gamma(\beta+\cos\theta')}\tag{1}.$$
We can compute $\tan{\frac{\theta}{2}}$ using the trigonometric identity
$$\tan{\frac{\theta}{2}}=\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\tag{2}$$
if we know $\cos\theta$, and we can compute $\cos\theta$ from $\tan\theta$ using another trigonometric identity,
$$\cos\theta=\frac{1}{\sqrt{1+\tan^2{\theta}}}\tag{3}.$$
Substituting (1) into (3) gives
$$\begin{align}\cos\theta&=\frac{1}{\sqrt{1+\left(\frac{\sin\theta'}{\gamma(\beta+\cos\theta')}\right)^2}}\\
&=\frac{\beta+\cos\theta'}{\sqrt{(\beta+\cos\theta')^2+(1-\beta^2)\sin^2\theta'}}\\
&=\frac{\beta+\cos\theta'}{1+\beta\cos\theta'}
\end{align}\tag{4}$$
and substituting (4) into (2) gives
$$\begin{align}\tan{\frac{\theta}{2}}&=\sqrt{\frac{1-\frac{\beta+\cos\theta'}{1+\beta\cos\theta'}}{1+\frac{\beta+\cos\theta'}{1+\beta\cos\theta'}}}\\
&=\sqrt{\frac{1+\beta\cos\theta'-\beta-\cos\theta'}{1+\beta\cos\theta'+\beta+\cos\theta'}}\\
&=\sqrt{\frac{(1-\beta)(1-\cos\theta')}{(1+\beta)(1+\cos\theta')}}\\
&=\sqrt{\frac{1-\beta}{1+\beta}}\tan{\frac{\theta'}{2}}
\end{align}\tag{5}.$$
Expressing $\beta$ in terms of $\xi$, we have
$$\begin{align}\sqrt{\frac{1-\beta}{1+\beta}}&=\sqrt{\frac{1-\tanh\xi}{1+\tanh\xi}}\\
&=\sqrt{\frac{\cosh\xi-\sinh\xi}{\cosh\xi+\sinh\xi}}\\
&=\sqrt{\frac{e^{-\xi}}{e^\xi}}\\
&=e^{-\xi}
\end{align}\tag{6}.$$
Substituting (6) into (5), we get
$$\tan{\frac{\theta}{2}}=e^{-\xi}\tan{\frac{\theta'}{2}}\tag{7},$$
which is the reverse of what you stated. For positive $\beta$, according to (1), $\theta$ should be less than $\theta'$.
