Question

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

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

Answer 1

Answer: (D)

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