Question

If $\sin \mathrm{\theta }=\frac{\sqrt{3}}{2}\mathrm{then}\left(\mathrm{cosec}\mathrm{\theta }+\cot \mathrm{\theta }\right)=?$

(a) $\left(2+\sqrt{3}\right)$

(b) $2\sqrt{3}$

(c) $\sqrt{2}$

(d) $\sqrt{3}$

Answer 1

$$\begin{gathered} \mathrm{Given}:\sin \theta =\frac{\sqrt{3}}{2} \\ \mathrm{Since},\sin \theta =\frac{P}{H} \\ \Rightarrow P=\sqrt{3}\mathrm{and}H=2 \\ \mathrm{Using}\mathrm{Pythagoras}\mathrm{theorem}, \\ {P}^2+B^2=H^2 \\ \Rightarrow {\left(\sqrt{3}\right)}^2+B^2=2^2 \\ \Rightarrow B^2=4-3 \\ \Rightarrow B^2=1 \\ \Rightarrow B=1 \\ \mathrm{Therefore}, \\ \mathrm{cosec}\theta =\frac{H}{P}=\frac{2}{\sqrt{3}} \\ \cot \theta =\frac{B}{P}=\frac{1}{\sqrt{3}} \\ \mathrm{Now}, \\ \left(\mathrm{cosec}\theta +\cot \theta \right)=\left(\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}\right) \\ =\left(\frac{2+1}{\sqrt{3}}\right) \\ =\left(\frac{3}{\sqrt{3}}\right) \\ =\sqrt{3} \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{d}\right). \end{gathered}$$