Question:
If $\tan \theta=\sqrt{3}$ then $\sec \theta=$ ?
(a) 2
(b) $\frac{1}{2}$
(c) $\frac{\sqrt{3}}{2}$
(d) $\frac{2}{\sqrt{3}}$
Solution:
Given : $\tan \theta=\frac{\sqrt{3}}{1}$
Since, $\tan \theta=\frac{P}{B}$
$\Rightarrow P=\sqrt{3}$ and $B=1$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow(\sqrt{3})^{2}+1^{2}=H^{2}$
$\Rightarrow H^{2}=3+1$
$\Rightarrow H^{2}=4$
$\Rightarrow H=2$
Therefore,
$\sec \theta=\frac{H}{B}=\frac{2}{1}=2$
Hence, the correct option is (a).
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