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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

$2 \sqrt{6}$

Vedclass · Answer

(C) $\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 :

Similar Questions

If $\operatorname{cosec} \theta - \cot \theta = 2017$,then the quadrant in which $\theta$ lies is

Which of the following relations is correct?

Find the values of the other five trigonometric functions if $\cot x = \frac{3}{4}$ and $x$ lies in the third quadrant.

If $\sin (\alpha - \beta ) = \frac{1}{2}$ and $\cos (\alpha + \beta ) = \frac{1}{2},$ where $\alpha$ and $\beta$ are positive acute angles,then:

If $\cot x = -\frac{5}{12}$ and $x$ lies in the second quadrant,find the values of the other five trigonometric functions.

Vedclass Test Series

Exam Paper Generator

Online Exam Module