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

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Question

(B) Let $x = \cot 70^{\circ} + 4 \cos 70^{\circ}$.We can write this as $x = \frac{\cos 70^{\circ}}{\sin 70^{\circ}} + 4 \cos 70^{\circ} = \frac{\cos 7

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(B) Let $x = \cot 70^{\circ} + 4 \cos 70^{\circ}$.We can write this as $x = \frac{\cos 70^{\circ}}{\sin 70^{\circ}} + 4 \cos 70^{\circ} = \frac{\cos 70^{\circ} + 4 \sin 70^{\circ} \cos 70^{\circ}}{\sin 70^{\circ}}$.Using the identity $2 \sin 	heta \cos 	heta = \sin 2	heta$,we get $x = \frac{\cos 70^{\circ} + 2 \sin 140^{\circ}}{\sin 70^{\circ}}$.Since $\sin 140^{\circ} = \sin(180^{\circ} - 40^{\circ}) = \sin 40^{\circ}$,we have $x = \frac{\cos 70^{\circ} + 2 \sin 40^{\circ}}{\sin 70^{\circ}}$.Using $\cos 70^{\circ} = \sin 20^{\circ}$,$x = \frac{\sin 20^{\circ} + 2 \sin 40^{\circ}}{\sin 70^{\circ}} = \frac{\sin 20^{\circ} + \sin 40^{\circ} + \sin 40^{\circ}}{\sin 70^{\circ}}$.Using $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$,we get $\sin 20^{\circ} + \sin 40^{\circ} = 2 \sin 30^{\circ} \cos 10^{\circ} = 2 	imes \frac{1}{2} 	imes \cos 10^{\circ} = \cos 10^{\circ}$.So,$x = \frac{\cos 10^{\circ} + \sin 40^{\circ}}{\sin 70^{\circ}} = \frac{\sin 80^{\circ} + \sin 40^{\circ}}{\sin 70^{\circ}}$.Using $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$,we get $x = \frac{2 \sin 60^{\circ} \cos 20^{\circ}}{\sin 70^{\circ}} = \frac{2 	imes \frac{\sqrt{3}}{2} 	imes \cos 20^{\circ}}{\cos 20^{\circ}} = \sqrt{3}$.

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

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