Question

The value of $\frac{\sqrt{3}}{\sin15^{\circ}} - \frac{\sqrt{1}}{\cos15^{\circ}}$ 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)

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