A projectile can have the same range $$'R\,'$$ for two angles of projection. If $$'{T_1}'$$  and $$'{T_2}'$$  to be time of flights in the two cases, then the product of the two time of flights is directly proportional to.

Solution :
The angle for which the ranges are same is complementary.
Let one angle be $$\theta ,$$ then other is $${90^ \circ } - \theta $$
$$\eqalign{ & {T_1} = \frac{{2u\,\sin \theta }}{g},\,\,\,{T_2} = \frac{{2u\,\cos \theta }}{g} \cr & {T_1}{T_2} = \frac{{4{u^2}\,\sin \theta \cos \theta }}{g} = 2R\,\,\,\left( {\because R = \frac{{{u^2}{{\sin }^2}\theta }}{g}} \right) \cr} $$
Hence it is proportional to $$R.$$