cosec⁡15∘+sec⁡15∘ is equal to

cosec⁡15∘+sec⁡15∘=2sin⁡15∘+cos⁡15∘2sin⁡15∘cos⁡15∘2sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘+cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘sin⁡3∘=4322−122+322+122=42322=26

cosec⁡15∘+sec⁡15∘ is equal to

cosec⁡15∘+sec⁡15∘=2sin⁡15∘+cos⁡15∘2sin⁡15∘cos⁡15∘2sin⁡45∘cos⁡30∘−cos⁡45∘sin⁡30∘+cos⁡45∘cos⁡30∘+sin⁡45∘sin⁡30∘sin⁡3∘=4322−122+322+122=42322=26

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

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$2 \sqrt{6}$

$2 \sqrt{2}$

$\sqrt{6}$

$\sqrt{6} + \sqrt{2}$

answer is C.

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Detailed Solution

$\begin{array}{l} cosec 15^{\circ} + \sec 15^{\circ} \\ = \frac{2 (\sin 15^{\circ} + \cos 15^{\circ})}{2 \sin 15^{\circ} \cos 15^{\circ}} \\ \frac{2 (\begin{matrix} \sin 45^{\circ} \cos 30^{\circ} - \cos 45^{\circ} \sin 30^{\circ} \\ + \cos 45^{\circ} \cos 30^{\circ} + \sin 45^{\circ} \sin 30^{\circ} \end{matrix})}{\sin 3^{\circ}} \\ = 4 [\frac{\sqrt{3}}{2 \sqrt{2}} - \frac{1}{2 \sqrt{2}} + \frac{\sqrt{3}}{2 \sqrt{2}} + \frac{1}{2 \sqrt{2}}] \\ = 4 [\frac{2 \sqrt{3}}{2 \sqrt{2}}] = 2 \sqrt{6} \end{array}$