Question

The value of $\cot \, 70° + 4 \cos\, 70° $ is

• A. $ 1 / \sqrt{3} $
• B. $ \sqrt{3} $
• C. $ 2 / \sqrt {3} $
• D. $ 1/2 $

Answer 1

Answer: (B)

Given, $\cot 70^{\circ}+4 \cos 70^{\circ}$
$=\frac{\cos 70^{\circ}}{\sin 70^{\circ}}+4 \cos 70^{\circ}$
$=\frac{\cos 70^{\circ}+4 \sin 70^{\circ} \cos 70^{\circ}}{\sin 70^{\circ}}$
$=\frac{\cos 70^{\circ}+2 \sin \left(180^{\circ}-40^{\circ}\right)}{\sin 70^{\circ}}$
$=\frac{\cos \left(90^{\circ}-20^{\circ}\right)+2 \sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{\sin 20^{\circ}+\sin 40^{\circ}+\sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{2 \sin 30^{\circ} \cos 10^{\circ}+\sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{2 \sin 30^{\circ} \cos \left(90^{\circ}-80^{\circ}\right)+\sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{\sin 80^{\circ}+\sin 40^{\circ}}{\sin 70^{\circ}}$
$=\frac{2 \sin 60^{\circ} \cos 20^{\circ}}{\sin 70^{\circ}}$
$=\frac{2\left(\frac{\sqrt{3}}{2}\right) \sin 70^{\circ}}{\sin 70^{\circ}}=\sqrt{3}$