$\begin{array}{1 1}R\\\large\frac{1}{R}\\\large\frac{1}{R^2}\\R^2\end{array} $

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Answer : R

$t_1=\large\frac{2v\sin \theta}{g}$

$t_2=\large\frac{2v\sin (90^{\large\circ}-\theta)}{g}=\frac{2v\cos\theta}{g}$

Now,$t_1t_2=\large\frac{(2v\sin\theta)(2v\cos\theta)}{g^2}$

$\Rightarrow \large\frac{2}{g}\bigg[\frac{v^2(2\sin \theta\cos\theta)}{g}\bigg]=\frac{2}{g}\frac{v^2\sin 2\theta}{g}=\frac{2}{g}=$$R$

$\therefore t_1t_2\propto R$

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