The value of frac{sin 300^{circ} tan 240^{cir | vedclass

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(C) We evaluate each trigonometric term using reduction formulas:$\sin 300^{\circ} = \sin(360^{\circ} - 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{

Vedclass · Accepted Answer

$\sqrt{6}$

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(C) We evaluate each trigonometric term using reduction formulas:$\sin 300^{\circ} = \sin(360^{\circ} - 60^{\circ}) = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}$$	an 240^{\circ} = 	an(180^{\circ} + 60^{\circ}) = 	an 60^{\circ} = \sqrt{3}$$\sec(-420^{\circ}) = \sec(420^{\circ}) = \sec(360^{\circ} + 60^{\circ}) = \sec 60^{\circ} = 2$$\cot(-315^{\circ}) = -\cot(315^{\circ}) = -\cot(360^{\circ} - 45^{\circ}) = -(-\cot 45^{\circ}) = 1$$\cos 210^{\circ} = \cos(180^{\circ} + 30^{\circ}) = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}$$\operatorname{cosec}(-315^{\circ}) = -\operatorname{cosec}(315^{\circ}) = -\operatorname{cosec}(360^{\circ} - 45^{\circ}) = -(-\operatorname{cosec} 45^{\circ}) = \sqrt{2}$Substituting these values into the expression:$= \frac{(-\frac{\sqrt{3}}{2}) \cdot (\sqrt{3}) \cdot (2)}{(1) \cdot (-\frac{\sqrt{3}}{2}) \cdot (\sqrt{2})} = \frac{-\sqrt{3} \cdot \sqrt{3}}{-\frac{\sqrt{3}}{2} \cdot \sqrt{2}} = \frac{-3}{-\frac{\sqrt{6}}{2}} = \frac{6}{\sqrt{6}} = \sqrt{6}$

The value of $\frac{\sin 300^{\circ} \tan 240^{\circ} \sec \left(-420^{\circ}\right)}{\cot \left(-315^{\circ}\right) \cos \left(210^{\circ}\right) \operatorname{cosec}\left(-315^{\circ}\right)}$ is

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