The value of text{cosec}10^{circ} - sqrt{3} t | vedclass

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Question

(A) Given expression: $\frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}}$$= \frac{\cos 10^{\circ} - \sqrt{3} \sin 10^{\circ}}{\sin 10^{\cir

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) Given expression: $\frac{1}{\sin 10^{\circ}} - \frac{\sqrt{3}}{\cos 10^{\circ}}$$= \frac{\cos 10^{\circ} - \sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$Multiply numerator and denominator by $2$:$= \frac{2(\frac{1}{2} \cos 10^{\circ} - \frac{\sqrt{3}}{2} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$$= \frac{2(\sin 30^{\circ} \cos 10^{\circ} - \cos 30^{\circ} \sin 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}}$Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$= \frac{2 \sin(30^{\circ} - 10^{\circ})}{\sin 10^{\circ} \cos 10^{\circ}} = \frac{2 \sin 20^{\circ}}{\sin 10^{\circ} \cos 10^{\circ}}$Using $\sin 2	heta = 2 \sin 	heta \cos 	heta$:$= \frac{2 \sin 20^{\circ}}{\frac{1}{2} \sin 20^{\circ}} = 4$

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

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