A projectile has the same range R for two angles of projections. If $T_{1}and{T}_2$ be the times of flight in the two cases, then (using as the angle of projection corresponding to $T_1$)

Horizontal range is same when angle of projection is $\theta and({90}^0-\theta )$

$\therefore R=\frac{u^2\sin 2\theta }{g}=\frac{2u^2\sin \theta \cos \theta }{g}$

When angle of projection is $\theta ,then$

$T_1=\frac{2u\sin \theta }{g}$

When angle of projection is $({90}^0-\theta ),then$

$T_2=\frac{2u\sin ({90}^0-\theta )}{g}=\frac{2u\cos \theta }{g}$

$\therefore T_l{T}_2=\frac{4u^2\sin \theta cos\theta }{g^2}=\left(\frac{2u^2\sin \theta \cos \theta }{g}\right)\left(\frac{2}{g}\right)=\frac{2R}{g}$

$\therefore T_1{T}_2\propto R$

$and\frac{T_1}{T_2}=\frac{2u\sin \theta /g}{2u\cos \theta /g}=\tan \theta$