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