The value of cot 70^{circ} + 4cos 70^{circ} i | vedclass

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(B) Let $E = \cot 70^{\circ} + 4\cos 70^{\circ} = \frac{\cos 70^{\circ} + 4\sin 70^{\circ}\cos 70^{\circ}}{\sin 70^{\circ}}$Using $2\sin 	heta \cos \

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$\sqrt{3}$

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(B) Let $E = \cot 70^{\circ} + 4\cos 70^{\circ} = \frac{\cos 70^{\circ} + 4\sin 70^{\circ}\cos 70^{\circ}}{\sin 70^{\circ}}$Using $2\sin 	heta \cos 	heta = \sin 2	heta$,we have $4\sin 70^{\circ}\cos 70^{\circ} = 2\sin 140^{\circ}$.$E = \frac{\cos 70^{\circ} + 2\sin 140^{\circ}}{\sin 70^{\circ}} = \frac{\sin 20^{\circ} + 2\sin(180^{\circ} - 40^{\circ})}{\sin 70^{\circ}}$$E = \frac{\sin 20^{\circ} + 2\sin 40^{\circ}}{\sin 70^{\circ}} = \frac{\sin 20^{\circ} + \sin 40^{\circ} + \sin 40^{\circ}}{\sin 70^{\circ}}$Using $\sin C + \sin D = 2\sin(\frac{C+D}{2})\cos(\frac{C-D}{2})$,we get $\sin 20^{\circ} + \sin 40^{\circ} = 2\sin 30^{\circ}\cos 10^{\circ} = \cos 10^{\circ}$.Since $\cos 10^{\circ} = \sin 80^{\circ}$,$E = \frac{\sin 80^{\circ} + \sin 40^{\circ}}{\sin 70^{\circ}}$$E = \frac{2\sin 60^{\circ}\cos 20^{\circ}}{\cos 20^{\circ}} = 2\sin 60^{\circ} = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}$.

The value of $\cot 70^{\circ} + 4\cos 70^{\circ}$ is

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