Question

The value of $\sqrt{3}\cot {20}^0-4\cos {20}^0$ is

• A. $1$
• B. $-1$
• C. $0$
• D. None of these

Answer 1

Answer: (A)

$1$

$\sqrt{3}\cot {20}^0-4\cos {20}^0$
$=\sqrt{3}\frac{\cos {20}^0}{\sin {20}^0}-4\cos {20}^0$
$=\frac{\sqrt{3}\cos {20}^0-4\sin {20}^0\cos {20}^0}{\sin {20}^0}$
$=\frac{2\cos {20}^0\sin {60}^0-\sin {40}^0}{\sin {20}^0}$
$=\frac{\sin {80}^0+\sin {40}^0-2\sin {40}^0}{\sin {20}^0}$
$=\frac{\sin {80}^0-\sin {40}^0}{\sin {20}^0}$
$=\frac{2\cos {60}^0\sin {20}^0}{\sin {20}^0}=1$