Question

The value of $\frac{\sqrt{3}}{\sin 15^{\circ }}-\frac{\sqrt{1}}{\cos 15^{\circ }}$ is equal to

• A. $4\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\sqrt{2}$
• D. $\frac{1}{\sqrt{2}}$
• E. $\frac{\sqrt{3}}{2}$

Answer 1

Answer: (A)

$\frac{\sqrt{3}}{\sin 15^{\circ }}-\frac{1}{\cos 15^{\circ }}=\frac{\sqrt{3}\cos 15^{\circ }-\sin 15^{\circ }}{\sin 15^{\circ }\cos 15^{\circ }}$
$=\frac{\frac{\sqrt{3}}{2}\cos 15^{\circ }-\frac{1}{2}\sin 15^{\circ }}{\frac{1}{2}\sin 15^{\circ }\cos 15^{\circ }}$
$=\frac{\sin 60^{\circ }\cos 15^{\circ }-\cos 60^{\circ }\sin 15^{\circ }}{\frac{1}{4}\left(\sin 30^{\circ }\right)}$
$=\frac{\sin \left(60^{\circ }-15^{\circ }\right)}{\frac{1}{4}\times \frac{1}{2}}=\frac{\sin 45^{\circ }}{\frac{1}{8}}=8\times \frac{1}{\sqrt{2}}=\frac{8\times \sqrt{2}}{2}=4\sqrt{2}$