how to prove that two angles made by unprimed axes are same in minkowski diagram I am reading special theory of relativity by resnick.In supplemntory topic A it is given that two inertial reference frame s and s' where s' is moving relative to s with uniform velocity.The diagram is shown below
It is given that if X' and ct' makes the same angle with x and ct respectively.
How to prove that two angles are same?
 A: The primed coordinates are a hyperbolic rotation from the unprimed using
 $$\begin{bmatrix}ct'\\x'\end{bmatrix}=\begin{bmatrix}\cosh\delta&-\sinh\delta\\-\sinh\delta&\cosh\delta\end{bmatrix}\begin{bmatrix}ct\\x\end{bmatrix}.$$ This rotation arises from the invariance of the 4-vector length
$$(ct)^2-x^2=(ct')^2-x'^2.$$
We can find the points for the $ct'$ axis in terms of the $ct, x$ coordinates by using the inverse rotation (drop the minus signs from the $-\sinh\delta$ terms) by setting $x'=0$:
$$\begin{bmatrix}ct\\x\end{bmatrix}=\begin{bmatrix}\cosh\delta&\sinh\delta\\\sinh\delta&\cosh\delta\end{bmatrix}\begin{bmatrix}ct'\\0\end{bmatrix}=\begin{bmatrix}ct'\cosh\delta\\ct'\sinh\delta\end{bmatrix}.$$
The ratio of ct/x for this set of points is a constant, $\coth\delta$, and this is the slope of the $ct'$ axis.
If you do the same process for the $(ct,x)$ coordinates of the $x'$ axis, you should find that the ratio if $\tanh\delta$. This is the slope of the $x'$ axis and is the reciprocal of the $\coth$ term.
In terms of he speed of the primed system relative to the unprimed we have $\beta = \frac{\sinh\delta}{\cosh\delta}=\tanh\delta$. The angle that the primed axes make with their respective unprimed is $\alpha=\arctan\beta.$
EDIT: Why are we interested in rotations? Because when we change coordinate systems we expect certain quantities, such as vectors, to have invariant lengths.
3D vectors are rotated by using trigonometric functions. 
 $$\begin{bmatrix}x'\\y'\end{bmatrix}=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}.$$ Depending on whether you are rotating the vector or the coordinate system, the signs of the $\sin\theta$ terms may be reversed, but that's not important for this discussion. What is important: the length of the vector cannot change when the coordinate system used to describe the vector rotates. You should already know this. If you don't, there are several web resources, including Wikipedia.
In 4-space, we expect the spacetime coordinates of an event to rotate with a constant difference between the space and coordinates, $(ct)^2−x^2=(ct′)^2−x′^2$. But if $(ct)^2−x^2$ is constant, that defines a hyperbola. The rotation of a hyperbola to a new coordinate basis involves hyperbolic functions. see this website (not mine).
Thomas Moore's book, A Traveler's Guide to Spacetime or Spacetime Physics by Resnick and Wheeler will talk about this. If you are unfamiliar with using matrices and doing coordinate rotations, 2nd year is not too early to learn about it. Doing Lorentz transformations with matrices is MUCH easier than memorizing those crazy formulas.  
The relationships are $\cosh\delta =\gamma$, $\sinh\delta = \gamma\beta$, and $\beta = \tanh\delta$. The $\delta$ term is, in simple terms, the sum of infinitesimal changes in $\beta$ and it's called the rapidity or the boost. For example, if you observe Object A moving at $\beta_A=0.5$ toward you, and Object A throws Object B toward you with $\beta_B=0.6$ relative to B, what will you measure for the speed of B toward you? Let's find the two $\delta$ terms:
$$\delta_A=\text{arctanh(.5)=0.5493 and }\delta_B=\text{ arctanh(0.6)=0.6931.}$$
So the total rapidity in your frame is 0.5493+0.6931=1.2424. Next, we find your observation of object B:$$\beta = \tanh(1.2424)=0.846$$. Check this problem out with the standard formulas that most SR classes make you memorize. Also notice that no matter how big $\delta$ becomes, $\beta < 1$. 
Hyperbolic rotation of coordinates in spacetime gives us an asymptotic speed limit in spacetime.
