Question

Exact value of $\sqrt{3}\cot {20}^o-4\cos {20}^o$ is-

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

Answer 1

The correct option is B $1$

We know that,

$\sin ({60}^0-{20}^0)=\sin {40}^0$

$\Rightarrow \sin {60}^0\cos {20}^0-\cos {60}^0\sin {20}^0=\sin {40}^0$

$\Rightarrow \frac{\sqrt{3}}{2}\cos {20}^0-\frac{1}{2}\sin {20}^0=2\sin {20}^0\cos {20}^0$

$\Rightarrow \frac{\sqrt{3}}{2}\cos {20}^0-2\sin {20}^0\cos {20}^0=\frac{1}{2}\sin {20}^0$

$\Rightarrow 2\frac{\sqrt{3}\cos {20}^0}{2\sin {20}^0}-\frac{4\sin {20}^0\cos {20}^0}{\sin {20}^0}=1$

$\Rightarrow \sqrt{3}\cot {20}^0-4\cos {20}^0=1$