frac{1}{{sin 10^circ }} - frac{{sqrt 3 }}{{co | vedclass

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Question

(D) Given expression: $\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$Taking the common denominator: $\frac{\cos 10^\circ - \sqrt{3} \sin 10

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(D) Given expression: $\frac{1}{\sin 10^\circ} - \frac{\sqrt{3}}{\cos 10^\circ}$Taking the common denominator: $\frac{\cos 10^\circ - \sqrt{3} \sin 10^\circ}{\sin 10^\circ \cos 10^\circ}$Multiply the numerator and denominator by $2$: $\frac{2(\frac{1}{2} \cos 10^\circ - \frac{\sqrt{3}}{2} \sin 10^\circ)}{\sin 10^\circ \cos 10^\circ}$Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$,we rewrite the numerator as $2 \sin(30^\circ - 10^\circ) = 2 \sin 20^\circ$.For the denominator,use the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,so $\sin 10^\circ \cos 10^\circ = \frac{1}{2} \sin 20^\circ$.Substituting these back: $\frac{2 \sin 20^\circ}{\frac{1}{2} \sin 20^\circ} = 4 	imes \frac{\sin 20^\circ}{\sin 20^\circ} = 4$.

$\frac{1}{{\sin 10^\circ }} - \frac{{\sqrt 3 }}{{\cos 10^\circ }} = $

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