Let the angle of projection in the two cases by $\theta_0$ and $(90^{\circ}- \theta_{0})$ i.e $v_0 \sin \theta$ in the second case. Hence,

$R= (v_0 \cos \theta_0)t_1$

$R= (v_0 \sin \theta_0) t_2$

therefore $t_1t_2 =\large\frac{R^2}{v_0 ^2 \sin \theta _0 \cos \theta_0 }$----(i)

But $R= \large\frac{2v_0^2 \sin \theta_0 \cos \theta_0}{g}$-----(ii)

Hence from (i) and (ii)

$t_1t_2=\large\frac{2R}{g}$