Explanation

(a)
Let the equation of rectangular hyperbola is $xy = c^{2}.$
Take any four points on the hyperbola
$P\left( ct_{1},\frac{c}{t_{1}} \right),\ Q\left( ct_{2},\frac{c}{t_{2}} \right),\ R\left( ct_{3},\frac{c}{t_{3}} \right)\ and\ S\left( ct_{4},\frac{c}{t_{4}} \right)$ Such that $PQ$ is perpendicular to$RS$.
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Since, $OP$ makes angle $\alpha$ with$OX$.
Therefore, $\tan{\alpha = \frac{\frac{c}{t_{1}}}{ct_{1}} = \frac{1}{t_{1}^{2}}}$
Similarly, $\tan{\beta = \frac{1}{t_{2}^{2}},\tan{\gamma = \frac{1}{t_{3\ }^{2}}}}$ and $\tan{\delta = \frac{1}{t_{4}^{2}}}$
$\therefore\ \tan{\alpha\tan{\beta\tan{\gamma\tan{\delta = \frac{1}{t_{1}^{2}t_{2}^{2}t_{3}^{2}t_{4}^{2}}}}}}$ …(i)
Now, $PQ$ is perpendicular to $RS$.
$\therefore\ \ \frac{\frac{c}{t_{2}} - \frac{c}{t_{1}}}{ct_{2} - c_{1}} \times \frac{\frac{c}{t_{4}} - \frac{c}{t_{3}}}{ct_{4} - ct_{3}} = - 1$
$\Rightarrow - \frac{1}{t_{1}t_{2}} \times \left( - \frac{1}{t_{3}t_{4}} \right) = - 1$
$\Rightarrow \frac{1}{t_{1}t_{2}t_{3}t_{4}} = - 1$
$\Rightarrow t_{1}t_{2}t_{3}t_{4} = - 1$
From Eq.(i),
$\tan{\alpha\tan{\beta\tan{\gamma\tan{\delta = 1}}}}$