1. Use conservation of linear momentum

Since the collision is completely inelastic, the two particles stick together and move as a single particle. The direction of motion of the combined particle is the direction of the vector sum of the initial momenta.

Because both particles have equal mass $M$ and equal speed $v$, each has momentum magnitude

$p = Mv$

So we only need to add their momentum vectors.

2. Resolve the two momenta into components

From the figure, particle $A$ and particle $B$ move with speed $v$ along two given directions. Let their momentum components along $x$ and $y$ be added.

The resultant direction satisfies

$\tan\theta = \frac{P_y}{P_x}$

where $P_x$ and $P_y$ are the net momentum components.

Using the directions shown in the figure, the summed components come out to be proportional to

$P_y \propto \sqrt{3}+\sqrt{2}, \qquad P_x \propto 1-\sqrt{2}$

Hence,

$\tan\theta = \frac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}$

3. Match with the options

This corresponds to:

$\boxed{\text{A}}$

4. Final answer

$\boxed{\tan\theta=\frac{\sqrt{3}+\sqrt{2}}{1-\sqrt{2}}}$