In the given figure, $A B$ is the diameter of a circle. Find
$t a n A \times t a n B$ .

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Solution

We know that for any angle θ,
$t a n θ = \frac{P e r p e n d i c u l a r}{B a s e}$

For $t a n A$ ,
perpendicular = $2 c m$ and base = $3 c m$ .
Therefore, $t a n A = \frac{2}{3}$ (0.5 Marks)

Similarly,
for $t a n B$ , perpendicular = $3 c m$ and base = $2 c m$ .
Therefore,
$t a n B = \frac{3}{2}$ (0.5 Marks)

Now,
$t a n A \times t a n B = \frac{2}{3} \times \frac{3}{2}$
$= 1$ (1 Mark)

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