First slide

Multiple and sub- multiple Angles

Question

The value of $\cot 36^{\circ} \cot 72^{\circ}$ is

Moderate

A

$1 / 5$
B

$1 / \sqrt{5}$
C

1
D

$1 / 3$

Solution

We find that
$\begin{array}{l} \cot^{2} 36^{\circ} \cot^{2} 72^{\circ} \\ = \frac{\cos^{2} 36^{\circ} \cos^{2} 72^{\circ}}{\sin^{2} 36^{\circ} \sin^{2} 72^{\circ}} \\ = \frac{(1 + \cos 72^{\circ}) (1 + \cos 144^{\circ})}{(1 - \cos 72^{\circ}) (1 - \cos 144^{\circ})} \\ = \frac{(1 + \cos 72^{\circ}) (1 - \cos 36^{\circ})}{(1 - \cos 72^{\circ}) (1 + \cos 36^{\circ})} \\ = \frac{1 + \cos 72^{\circ} - \cos 36^{\circ} - \cos 72^{\circ} \cos 36^{\circ}}{1 - \cos 72^{\circ} + \cos 36^{\circ} - \cos 72^{\circ} \cos 36^{\circ}} \\ = \frac{1 - (\cos 36^{\circ} - \cos 72^{\circ}) - \cos 72^{\circ} \cos 36^{\circ}}{1 + \cos 36^{\circ} - \cos 72^{\circ} - \cos 72^{\circ} \cos 36^{\circ}} \end{array}$
$= \frac{1 - \frac{1}{2} - \frac{1}{4}}{1 + \frac{1}{2} - \frac{1}{4}} = \frac{1}{5}$ $[\begin{array}{l} ∵ \cos 36^{\circ} - \cos 72^{\circ} = \frac{1}{2} \\ and, \cos 36^{\circ} \cos 72^{\circ} = \frac{1}{4} \end{array}]$
$∴ \cot^{2} 36^{\circ} \cot^{2} 72^{\circ} = \frac{1}{5} \Rightarrow \cot 36^{\circ} \cot 72^{\circ} = \frac{1}{\sqrt{5}}$

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