Calculate without using trigonometric tables $4 cos 20^{o} - \sqrt{3} cot 20^{o}$ .

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Solution

We know that,

$\Rightarrow sin 2 A = 2 sin A sin B$

So, $sin (60^{0} - 20^{0}) = sin 40^{0} = 2 sin 20^{0} cos 20^{0}$

$\Rightarrow sin (A - B) = sin A cos B - cos A sin B$

$\Rightarrow sin 60^{0} cos 20^{0} - cos 60^{0} sin 20^{0} = 2 sin 20^{0} cos 20^{0}$

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

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

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

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

Therefore, $\Rightarrow 4 cos 20^{0} - \sqrt{3} c o t 20^{0} = - 1$

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