The value of frac{sqrt{3}text{cosec } 20^{cir | vedclass

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

Question

(C) Let $E = \frac{\sqrt{3}	ext{cosec } 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}}$.First, simplify th

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(C) Let $E = \frac{\sqrt{3}	ext{cosec } 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}}$.First, simplify the numerator: $\sqrt{3}	ext{cosec } 20^{\circ}-\sec 20^{\circ} = \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \frac{\sqrt{3}\cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2(\frac{\sqrt{3}}{2}\cos 20^{\circ} - \frac{1}{2}\sin 20^{\circ})}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2\sin(60^{\circ}-20^{\circ})}{\sin 20^{\circ}\cos 20^{\circ}} = \frac{2\sin 40^{\circ}}{\sin 20^{\circ}\cos 20^{\circ}}$.Next, simplify the denominator: $\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ} = \cos 20^{\circ}\cos 40^{\circ} \cdot \frac{1}{2} \cdot \cos 80^{\circ} = \frac{1}{2} \cdot \frac{\sin(2^3 \cdot 20^{\circ})}{2^3 \sin 20^{\circ}} = \frac{1}{16} \cdot \frac{\sin 160^{\circ}}{\sin 20^{\circ}} = \frac{1}{16} \cdot \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = \frac{1}{16}$.Thus, $E = \frac{2\sin 40^{\circ} / (\sin 20^{\circ}\cos 20^{\circ})}{1/16} = \frac{4\sin 40^{\circ} / \sin 40^{\circ}}{1/16} = 4 	imes 16 = 64$.

The value of $\frac{\sqrt{3}\text{cosec } 20^{\circ}-\sec 20^{\circ}}{\cos 20^{\circ}\cos 40^{\circ}\cos 60^{\circ}\cos 80^{\circ}}$ is equal to

Similar Questions

The value of $\text{cosec}10^{\circ} - \sqrt{3} \text{sec}10^{\circ}$ is equal to:

Evaluate: $\sin 6^{\circ} + \sin 54^{\circ} + \sin 126^{\circ} + \cos 156^{\circ}$

If $\alpha, \beta$ are acute angles such that $\sin \beta=2 \sin \alpha$ and $3 \cos \beta=2 \cos \alpha$,then $\sec (\alpha+\beta)=$

$\left( {1 + \cos \frac{\pi }{8}} \right)\,\left( {1 + \cos \frac{{3\pi }}{8}} \right)\,\left( {1 + \cos \frac{{5\pi }}{8}} \right)\,\left( {1 + \cos \frac{{7\pi }}{8}} \right) = $

It is known that $\sin \beta = \frac{4}{5}$ and $0 < \beta < \pi$. Then the value of $\frac{\sqrt{3} \sin(\alpha + \beta) - \frac{2}{\cos(\pi/6)} \cos(\alpha + \beta)}{\sin \alpha}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module