There's loads of ugly algebraic manipulation involved in this question, so apologies if I don't type out full working in getting from one stage to the next.
Considering horizontal motion:
= horizontal displacement after time t
Considering vertical motion:
, because the particle just clears the walls
rearranging, we obtain
which rearranges to give
substituting this into (*), and using the identity
, we obtain
which rearranges to give the required result.
From our expression for tan(alpha), we get
Using the formula for tan(A+B) and rearranging:
is small, we can disregard its term in the denominator, giving the required result.
Now we similarly obtain:
replacing the part of the last fraction that equals 2V^2/g and rearranging:
Now the LHS =
, of which we can disregard the
term because it is very small. Therefore LHS
From this, we can conclude that
has the same sign as the RHS of equation (A). Therefore,
This rearranges to
we can disregard the
term because it is much smaller than all the others, and then we can square root to give the required result. The other result can easily be shown by reversing the > sign.