For a given velocity, a projectile has the same range R for two angles of projection if t1 and t2 arc the times of flight in the two cases then
t1t2∝R2
t1t2∝R
t1t2∝1R
t1t2∝1R2
For same range angle of projection should be θ and, 90-θ
So, time of flights t1=2usinθg and
t2=2usin(90−θ)g=2ucosθg
By multiplying t1t2=4u2sinθcosθg2
t1t2=2gu2sin2θg=2Rg⇒t1t2∝R