sqrt{3} operatorname{cosec} 20^{circ} - sec 2 | vedclass

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Question

(D) We have the expression: $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$ $= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) We have the 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 the numerator and denominator by $2$: $= \frac{2(\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ})}{\sin 20^{\circ} \cos 20^{\circ}}$ $= \frac{2(\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}}$ $= \frac{4 \sin(60^{\circ} - 20^{\circ})}{\sin 40^{\circ}}$ $= \frac{4 \sin 40^{\circ}}{\sin 40^{\circ}} = 4$

List-$I$	List-$II$
$A$. The period of $\sin^2 x$ is	$I$. $\frac{2\pi}{3}$
$B$. Maximum value of $\frac{\pi}{3}(\sqrt{3}\cos 3x + \sin 3x)$	$II$. $12\pi$
$C$. The period of $\sin \frac{x}{3} + \cos \frac{x}{2}$ is	$III$. $\frac{\pi}{2}$
$D$. Intersection points of $y=\|\sin x\|$ and $y=1$ in $(0, \pi)$	$IV$. $\frac{3\pi}{2}$
	$V$. $\pi$

List-$A$	List-$B$
$(I)$ $\tan \left(\frac{\alpha + \beta}{2}\right) =$	$(a)$ $\frac{b}{a}$
$(II)$ $\cos (\alpha + \beta) =$	$(b)$ $\frac{2ab}{a^2 + b^2}$
$(III)$ $\sin (\alpha + \beta) =$	$(c)$ $\frac{2ab}{a^2 - b^2}$
$(IV)$ $\tan (\alpha + \beta) =$	$(d)$ $\frac{a^2 - b^2}{a^2 + b^2}$

$\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$ is equal to

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