Question

The value of $\cot {70}^0+4\cos {70}^0$ is

• A. $\sqrt{2}$
• B. $\sqrt{3}$
• C. $0$
• D. $1$

Answer 1

The correct option is B $\sqrt{3}$

$\cot 70°+4\cos 70°$

$=\cot (90°-20°)+4\cos (90°-20°)$

$=\tan 20°+4\sin 20°$

$=\frac{\sin 20°}{\cos 20°}+2\times 2\sin 20°$

$=\frac{\sin 20°+2\times 2\sin 20°\cos 20°}{\cos 20°}$

$=\frac{\sin 20°+2\sin 40°}{\cos 20°}$

$=\frac{\sin 20°+\sin 40°+\sin 40°}{\cos 20°}$

$=\frac{2\sin 30°\cos 10°+\sin 40°}{\cos 20°}$

$=\frac{\cos (90°-80°)+\sin 40°}{\cos 20°}$

$=\frac{\sin 80°+\sin 40°}{\cos 20°}$

$=\frac{2\sin 60°\cos 20°}{\cos 20°}$

$=\frac{2\sqrt{3}}{2}$

$=\sqrt{3}$