Question

If $\mathrm{cosec}\mathrm{\theta }=\sqrt{10}\mathrm{then}\sec \mathrm{\theta }=?$

(a) $\frac{1}{\sqrt{1}0}$

(b) $\frac{2}{\sqrt{10}}$

(c) $\frac{3}{\sqrt{10}}$

(d) $\frac{\sqrt{10}}{3}$

Answer 1

$$\begin{gathered} \mathrm{Given}:\mathrm{cosec}\theta =\frac{\sqrt{10}}{1} \\ \mathrm{Since},\mathrm{cosec}\theta =\frac{H}{P} \\ \Rightarrow P=1\mathrm{and}H=\sqrt{10} \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow 1^2+B^2={\left(\sqrt{10}\right)}^2 \\ \Rightarrow B^2=10-1 \\ \Rightarrow B^2=9 \\ \Rightarrow B=3 \\ \mathrm{Therefore}, \\ \sec \theta =\frac{H}{B}=\frac{\sqrt{10}}{3} \end{gathered}$$

Hence, the correct option is (d).