2 sin ^{2} 30^{circ} cot 30^{circ}-3 cos ^{2} | vedclass

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Question

(B) Given expression: $2 \sin ^{2} 30^{\circ} \cot 30^{\circ}-3 \cos ^{2} 60^{\circ} \sec ^{2} 30^{\circ}$Substitute the trigonometric values: $\sin 3

Vedclass · Accepted Answer

$\frac{\sqrt{3}-2}{2}$

Vedclass · Answer

(B) Given expression: $2 \sin ^{2} 30^{\circ} \cot 30^{\circ}-3 \cos ^{2} 60^{\circ} \sec ^{2} 30^{\circ}$Substitute the trigonometric values: $\sin 30^{\circ} = \frac{1}{2}$,$\cot 30^{\circ} = \sqrt{3}$,$\cos 60^{\circ} = \frac{1}{2}$,$\sec 30^{\circ} = \frac{2}{\sqrt{3}}$.$= 2 \left(\frac{1}{2}ight)^{2} (\sqrt{3}) - 3 \left(\frac{1}{2}ight)^{2} \left(\frac{2}{\sqrt{3}}ight)^{2}$$= 2 	imes \frac{1}{4} 	imes \sqrt{3} - 3 	imes \frac{1}{4} 	imes \frac{4}{3}$$= \frac{\sqrt{3}}{2} - 1$$= \frac{\sqrt{3}-2}{2}$

$2 \sin ^{2} 30^{\circ} \cot 30^{\circ}-3 \cos ^{2} 60^{\circ} \sec ^{2} 30^{\circ} = \dots$

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