Question:
If $\operatorname{cosec} \theta=\sqrt{10}$ then $\sec \theta=?$
(a) $\frac{1}{\sqrt{10}}$
(b) $\frac{2}{\sqrt{10}}$
(c) $\frac{3}{\sqrt{10}}$
(d) $\frac{\sqrt{10}}{3}$
Solution:
Given : $\operatorname{cosec} \theta=\frac{\sqrt{10}}{1}$
Since, $\operatorname{cosec} \theta=\frac{H}{P}$
$\Rightarrow P=1$ and $H=\sqrt{10}$
Using Pythagoras theorem,
$P^{2}+B^{2}=H^{2}$
$\Rightarrow 1^{2}+B^{2}=(\sqrt{10})^{2}$
$\Rightarrow B^{2}=10-1$
$\Rightarrow B^{2}=9$
$\Rightarrow B=3$
Therefore,
$\sec \theta=\frac{H}{B}=\frac{\sqrt{10}}{3}$
Hence, the correct option is (d).
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