Relative velocity from space-time diagram Consider two frames of reference, with velocities V1 and V2 relative to an initial ground frame.  I have made the space-time diagrams for the three frames (Time represented on Y-axis).

As far as I know, the relative velocity between frame-1 and frame-2 (relative velocity is represented in green) is the projection of net velocity of frame-2 (i.e., t2 axis) on space-1 (x1) axis or Vise-Versa.
So,
  Considering $c = 1$,
$=> V21 = \sin(\beta-\alpha)$
$=> V21 = (\sin\beta \cos\alpha  - \cos\alpha \sin\beta)$ 
We can simply calculate values of sines and cosines.
$\sin\alpha = V1/1 = V1$
$\sin\beta = V2/1 = V2$ 
$=> V21 = ( V2\sqrt{(1-V1^2)} - V1\sqrt{(1-V2^2)})$ 
While we know the correct formula,
$V21 = (V2-V1)/(1-V1V2)$ 
What is wrong with this approach to find relative velocities?
It worked correctly for calculating the space-time transformation relations between two frames. 
Sorry for the bad diagram.
 A: The problem is that your spacetime diagram is wrong: you're using Euclidean geometry, when you should be using Minkowski. The actual diagram, if you'll forgive the low quality picture, looks something like this:

where I've drawn the $t^2 - x^2 = \pm 1$ hyperbolae, and the angles obey $\tan \alpha = v_1$ and $\tan \beta = v_2$. Note that that unit vectors have unit length with respect to the Minkowski metric but not with respect to the Euclidean metric: they lie on a hyperbola, not on a circle.
If you want to see how things look from frame 1, you have to move everything out along the hyperbolae:

where now $\delta$ is the unknown angle we want to find. To find it geometrically, however, we need to do some algebra first, because the geometry is hyperbolic, and it doesn't correspond to what happens when we draw on paper. And the key fact we need is that if given a velocity $v$ we define the rapidity $\eta$ by $v = \tanh \eta$, such that $\eta$ goes from zero to infinity as $v$ goes from zero to one, then when doing Lorentz transformations, rapidities simply add. This is still geometric, in a sense, because rapidity is a parameter along the hyperbolae, though it's not the usual arclength.
From this, the velocity addition formula is straightforward if we know our hyperbolic identities: since rapidities simply add and subtract, we have that $\eta_{21} = \eta_2-\eta_1$, and
$$v_{21} = \tanh(\eta_2 - \eta_1) = \frac{\tanh{\eta_2} - \tanh{\eta_1}}{1 - \tanh{\eta_2}\tanh{\eta_1}} = \frac{v_2 - v_1}{1 - v_1 v_2}.$$
