Question

$\operatorname{cosec} 15^{\circ}+\sec 15^{\circ}$ is equal to :

• A. $2 \sqrt{2}$
• B. $\sqrt{6}$
• C. $2 \sqrt{6}$
• D. $\sqrt{6}+\sqrt{2}$

Answer 1

Answer: (D)

$\text{cosec} \,15^{\circ}+\sec 15^{\circ}$
$=\frac{2\left(\sin 15^{\circ}+\cos 15^{\circ}\right)}{2 \sin 15^{\circ} \cos 15^{\circ}}$
$2\left(\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}\right.$
$=\frac{ \left.+\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}\right)}{\sin 30^{\circ}}$
$=4\left[\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right]$
$=4\left[\frac{\sqrt{3}+1}{2 \sqrt{2}}\right]$
$=(\sqrt{3}+1) \sqrt{2}=\sqrt{6}+\sqrt{2}$