First slide

Multiple and sub- multiple Angles

Question

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

Moderate

A

Cot 15^o
B

Tan 50^o
C

Cot 50^o
D

Cot $\sqrt{20 °}$

Solution

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

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