Question

If $\sin \theta =\frac{\sqrt{3}}{2}$, then $(\text{cosec}\theta +\cot \theta )=$

• A. $2\sqrt{3}$
• B. $\frac{2\sqrt{3}}{3}$
• C. $(2+\sqrt{3})$
• D. $\sqrt{3}$

Answer 1

The correct option is D $\sqrt{3}$

$\sin \theta =\frac{\sqrt{3}}{2}=\frac{BC}{AC}=\frac{\sqrt{3}x}{2x}\Rightarrow BC=\sqrt{3}x$ and $AC=2x$

By Pythagoras theorem, $A{B}^2=(A{C}^2-B{C}^2)$

$\therefore A{B}^2=(4x^2-3x^2)=x^2\Rightarrow AB=x$

$\text{cosec}\theta +\cot \theta =(\frac{AC}{BC}+\frac{AB}{BC})=\frac{2x}{\sqrt{3}x}+\frac{x}{\sqrt{3}x}=\frac{2+1}{\sqrt{3}}=\sqrt{3}$