The value of 3+cot⁡80∘cot⁡20∘cot⁡80∘+cot⁡20∘ is equal to

3+cos⁡80∘cos⁡20∘sin⁡80∘sin⁡20∘cos⁡80∘sin⁡80∘+cos⁡20∘sin⁡20∘=2sin⁡80∘sin⁡20∘+cos⁡80∘cos⁡20∘+sin⁡80∘sin⁡20∘sin⁡20∘cos⁡80∘+cos⁡20∘sin⁡80∘=−cos⁡100∘+cos⁡60∘+cos⁡60∘sin⁡100∘=1−cos⁡100∘sin⁡100=tan⁡50∘.

The value of 3+cot⁡80∘cot⁡20∘cot⁡80∘+cot⁡20∘ is equal to

3+cos⁡80∘cos⁡20∘sin⁡80∘sin⁡20∘cos⁡80∘sin⁡80∘+cos⁡20∘sin⁡20∘=2sin⁡80∘sin⁡20∘+cos⁡80∘cos⁡20∘+sin⁡80∘sin⁡20∘sin⁡20∘cos⁡80∘+cos⁡20∘sin⁡80∘=−cos⁡100∘+cos⁡60∘+cos⁡60∘sin⁡100∘=1−cos⁡100∘sin⁡100=tan⁡50∘.

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The value of $\frac{3 + \cot 80^{\circ} \cot 20^{\circ}}{\cot 80^{\circ} + \cot 20^{\circ}}$ is equal to

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cot 20°

cot 50°

$cot⁡ \sqrt{20^{\circ}}$

tan 50°

answer is B.

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Detailed Solution

$\begin{matrix} \frac{3 + \frac{\cos 80^{\circ} \cos 20^{\circ}}{\sin 80^{\circ} \sin 20^{\circ}}}{\frac{\cos 80^{\circ}}{\sin 80^{\circ}} + \frac{\cos 20^{\circ}}{\sin 20^{\circ}}} = \frac{2 \sin 80^{\circ} \sin 20^{\circ} + (\cos 80^{\circ} \cos 20^{\circ} + \sin 80^{\circ} \sin 20^{\circ})}{\sin 20^{\circ} \cos 80^{\circ} + \cos 20^{\circ} \sin 80^{\circ}} \\ = \frac{- \cos 100^{\circ} + \cos 60^{\circ} + \cos 60^{\circ}}{\sin 100^{\circ}} = \frac{1 - \cos 100^{\circ}}{\sin 100} = \tan 50^{\circ} . \end{matrix}$