The value of cos 1200^{circ} + tan 1485^{circ | vedclass

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Question

(A) Given expression: $\cos 1200^{\circ} + 	an 1485^{\circ}$ Using the property $\cos(n 	imes 360^{\circ} + 	heta) = \cos 	heta$ and $	an(n 	ime

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Given expression: $\cos 1200^{\circ} + 	an 1485^{\circ}$ Using the property $\cos(n 	imes 360^{\circ} + 	heta) = \cos 	heta$ and $	an(n 	imes 360^{\circ} + 	heta) = 	an 	heta$: $\cos 1200^{\circ} = \cos(3 	imes 360^{\circ} + 120^{\circ}) = \cos 120^{\circ}$ $	an 1485^{\circ} = 	an(4 	imes 360^{\circ} + 45^{\circ}) = 	an 45^{\circ}$ Now,calculate the values: $\cos 120^{\circ} = \cos(180^{\circ} - 60^{\circ}) = -\cos 60^{\circ} = -\frac{1}{2}$ $	an 45^{\circ} = 1$ Therefore,$\cos 1200^{\circ} + 	an 1485^{\circ} = -\frac{1}{2} + 1 = \frac{1}{2}$

The value of $\cos 1200^{\circ} + \tan 1485^{\circ}$ is

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