For a given velocity, a projectile has a same range R for two angles of projection, if  T₁and T₂ are the times of flight in the two cases then

For same range angle of projection should be $\mathrm{\theta }$ and  $90-\mathrm{\theta }$
  So, time of flights     ${\mathrm{T}}_1=\frac{2\mathrm{usin\theta }}{\mathrm{g}}\mathrm{and}$    ${\mathrm{T}}_2=\frac{2\mathrm{usin}(90-\mathrm{\theta })}{\mathrm{g}}=\frac{2\mathrm{ucos\theta }}{\mathrm{g}}$    By multiplying ${\mathrm{T}}_1{\mathrm{T}}_2=\frac{4{\mathrm{u}}^2\mathrm{sin\theta cos\theta }}{{\mathrm{g}}^2}$           ${\mathrm{T}}_1{\mathrm{T}}_2=\frac{2}{\mathrm{g}}\frac{\left({\mathrm{u}}^2\sin 2\mathrm{\theta }\right)}{\mathrm{g}}=\frac{2\mathrm{R}}{\mathrm{g}}\Rightarrow {\mathrm{T}}_1{\mathrm{T}}_2\propto \mathrm{R}$