operatorname{cosec} 15^{circ} + sec 15^{circ} | vedclass

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Question

(C) We have $\operatorname{cosec} 15^{\circ} + \sec 15^{\circ} = \frac{1}{\sin 15^{\circ}} + \frac{1}{\cos 15^{\circ}}$$= \frac{\cos 15^{\circ} + \sin

Vedclass · Accepted Answer

$2 \sqrt{6}$

Vedclass · Answer

(C) We have $\operatorname{cosec} 15^{\circ} + \sec 15^{\circ} = \frac{1}{\sin 15^{\circ}} + \frac{1}{\cos 15^{\circ}}$$= \frac{\cos 15^{\circ} + \sin 15^{\circ}}{\sin 15^{\circ} \cos 15^{\circ}}$Multiply numerator and denominator by $2$:$= \frac{2(\cos 15^{\circ} + \sin 15^{\circ})}{2 \sin 15^{\circ} \cos 15^{\circ}} = \frac{2(\cos 15^{\circ} + \sin 15^{\circ})}{\sin 30^{\circ}}$Since $\sin 30^{\circ} = \frac{1}{2}$,we have:$= 4(\cos 15^{\circ} + \sin 15^{\circ})$Using $\sin 15^{\circ} = \frac{\sqrt{6} - \sqrt{2}}{4}$ and $\cos 15^{\circ} = \frac{\sqrt{6} + \sqrt{2}}{4}$:$= 4 \left( \frac{\sqrt{6} + \sqrt{2}}{4} + \frac{\sqrt{6} - \sqrt{2}}{4} \right)$$= 4 \left( \frac{2 \sqrt{6}}{4} \right) = 2 \sqrt{6}$

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

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