If tan theta cdot cos 60^{circ} = frac{sqrt{3 | vedclass

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(D) Given: $	an 	heta \cdot \cos 60^{\circ} = \frac{\sqrt{3}}{2}$We know that $\cos 60^{\circ} = \frac{1}{2}$.Substituting this value: $	an 	heta

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$\frac{1}{\sqrt{2}}$

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(D) Given: $	an 	heta \cdot \cos 60^{\circ} = \frac{\sqrt{3}}{2}$We know that $\cos 60^{\circ} = \frac{1}{2}$.Substituting this value: $	an 	heta 	imes \frac{1}{2} = \frac{\sqrt{3}}{2}$Multiplying both sides by $2$,we get: $	an 	heta = \sqrt{3}$Since $	an 60^{\circ} = \sqrt{3}$,we have $	heta = 60^{\circ}$.Now,we need to find the value of $\sin (	heta - 15^{\circ})$.Substituting $	heta = 60^{\circ}$: $\sin (60^{\circ} - 15^{\circ}) = \sin 45^{\circ}$.Since $\sin 45^{\circ} = \frac{1}{\sqrt{2}}$,the final answer is $\frac{1}{\sqrt{2}}$.

If $\tan \theta \cdot \cos 60^{\circ} = \frac{\sqrt{3}}{2}$,then the value of $\sin (\theta - 15^{\circ})$ is

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