Answer
Please see the work below.
Work Step by Step
We know that the flight time of the first projectile is
$t=\frac{v_y1}{g}=\frac{vsin(45+\alpha)}{g}$
The flight time of the second projectile is given as
$t_a=\frac{v_y2}{g}=\frac{vsin(45-\alpha)}{g}$
Now we can find the ratio as
$\frac{t_1}{t_2}=\frac{\frac{vsin(45+\alpha)}{g}}{\frac{vsin(45-\alpha)}{g}}=\frac{vsin(45+\alpha)}{vsin(45-\alpha)}=tan(45+\alpha)$
