Q64 / 160 · TS EAMCET 2006 · Mathematics
$\operatorname{cosec} 15^{\circ}+\sec 15^{\circ}$ is equal to :
- A$2 \sqrt{2}$
- B$\sqrt{6}$
- C$2 \sqrt{6}$
- Correct answerD$\sqrt{6}+\sqrt{2}$
Solution
$\operatorname{cosec} 15^{\circ}+\sec 15^{\circ}$
$=\frac{2\left(\sin 15^{\circ}+\cos 15^{\circ}\right)}{2 \sin 15^{\circ} \cos 15^{\circ}}$
$=\frac{2\left(\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}+\cos 45^{\circ} \cos 30^{\circ}+\sin 45^{\circ} \sin 30^{\circ}\right)}{\sin 30^{\circ}}$
$=4\left[\frac{\sqrt{3}}{2 \sqrt{2}}-\frac{1}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}+\frac{1}{2 \sqrt{2}}\right]$
$=4\left\lceil\frac{\sqrt{3}+1}{2 \sqrt{2}}\right\rceil$
$=(\sqrt{3}+1) \sqrt{2}=\sqrt{6}+\sqrt{2}$
Hence, the correct answer is option: 'D'.