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$$\begin{gathered} \mathrm{Question} \mathrm{seems} \mathrm{incomplete}. \\ \mathrm{\alpha }+\mathrm{\beta }=9 \\ \mathrm{Take} \tan \\ \Rightarrow \tan \left(\mathrm{\alpha }+\mathrm{\beta }\right)=\tan \left(9\right) \\ \Rightarrow \frac{\mathrm{tan\alpha }+\mathrm{tan\beta }}{1-\mathrm{tan\alpha }.\mathrm{tan\beta }}=\tan \left(9\right) \\ \Rightarrow \frac{1}{\tan \left(9\right)}\left(\mathrm{tan\alpha }+\mathrm{tan\beta }\right)=1-\mathrm{tan\alpha }.\mathrm{tan\beta } \\ \Rightarrow \cot \left(9\right) \left(\mathrm{tan\alpha }+\mathrm{tan\beta }\right)=1-\mathrm{tan\alpha }.\mathrm{tan\beta } \\ \Rightarrow \mathrm{tan\alpha }.\mathrm{tan\beta }=1-\cot \left(9\right) \left(\mathrm{tan\alpha }+\mathrm{tan\beta }\right) \\ \mathrm{As} \cot \left(9\right) \mathrm{is} \mathrm{negative} \left[\mathrm{As} \frac{5\mathrm{\pi }}{2}<9<3\mathrm{\pi }\right] \\ \mathrm{tan\alpha }.\mathrm{tan\beta } \mathrm{when} \mathrm{tan\alpha }+\mathrm{tan\beta } \mathrm{is} \mathrm{minimum}. \\ \mathrm{Now} \mathrm{we} \mathrm{can} \mathrm{take} \mathrm{\alpha }=\frac{3\mathrm{\pi }}{2} \mathrm{and} \mathrm{\beta }=9-\frac{3\mathrm{\pi }}{2} \\ \mathrm{As} \tan \left(\frac{3\mathrm{\pi }}{2}\right)\to -\infty , \mathrm{hence} \mathrm{R}.\mathrm{H}.\mathrm{S}. \mathrm{will} \mathrm{tend} \mathrm{towards} -\infty . \\ \mathrm{Hence} \mathrm{minimum} \mathrm{value} \mathrm{of} \mathrm{tan\alpha }.\mathrm{tan\beta } \mathrm{will} \mathrm{tend} \mathrm{towards} -\infty \mathrm{if} \mathrm{\alpha }+\mathrm{\beta }=9 \end{gathered}$$

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