The value of the expression sqrt{3} operatorn | vedclass

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

Question

(C) Given expression: $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ $= \frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$ $= \fr

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Given expression: $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ $= \frac{\sqrt{3}}{\sin 20^{\circ}}-\frac{1}{\cos 20^{\circ}}$ $= \frac{\sqrt{3} \cos 20^{\circ}-\sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$ Multiply numerator and denominator by $2$: $= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ}-\frac{1}{2} \sin 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}}$ Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$ where $A=60^{\circ}$ and $B=20^{\circ}$: $= \frac{2(\sin 60^{\circ} \cos 20^{\circ}-\cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2}(2 \sin 20^{\circ} \cos 20^{\circ})}$ $= \frac{2 \sin(60^{\circ}-20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$

The value of the expression $\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}$ is equal to

Similar Questions

The expression $\cos^2 \phi + \cos^2(\theta + \phi) - 2 \cos \theta \cos \phi \cos(\theta + \phi)$ is

If $\frac{\sqrt{2} \sin \alpha}{\sqrt{1+\cos 2 \alpha}}=\frac{1}{7}$ and $\sqrt{\frac{1-\cos 2 \beta}{2}}=\frac{1}{\sqrt{10}}$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\tan (\alpha+2 \beta)$ is equal to

$\frac{\cos 15^{\circ} \cos^2 22\frac{1}{2}^{\circ} - \sin 75^{\circ} \sin^2 52\frac{1}{2}^{\circ}}{\cos^2 15^{\circ} - \cos^2 75^{\circ}} = $

If $\frac{\cos x}{\cos (x-2 y)}=\lambda$,then $\tan (x-y) \tan y$ is equal to

If $\sin (\theta-\alpha), \sin \theta$ and $\sin (\theta+\alpha)$ are in $H.P.$,then the value of $\cos 2 \theta$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module