Evaluate:
$ \frac{\sin 18^{\circ}}{\cos 72^{\circ}}+\sqrt{3}\left\{\tan 10^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right\} $

Given:

\( \frac{\sin 18^{\circ}}{\cos 72^{\circ}}+\sqrt{3}\left\{\tan 10^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right\} \).

To do:

We have to evaluate \( \frac{\sin 18^{\circ}}{\cos 72^{\circ}}+\sqrt{3}\left\{\tan 10^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right\} \). 

Solution:  

We know that,

$cos\ (90^{\circ}- \theta) = sin\ \theta$

$tan\ (90^{\circ}- \theta) = cot\ \theta$

$tan\ \theta \times \cot\ \theta=1$

Therefore,

$\frac{\sin 18^{\circ}}{\cos 72^{\circ}}+\sqrt{3}\left\{\tan 10^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right\}=\frac{\sin 18^{\circ}}{\cos( 90^{\circ}-18^{\circ})} +\sqrt{3}\left\{\tan 10^{\circ}\left(\frac{1}{\sqrt{3}}\right)\tan 40^{\circ}\tan( 90^{\circ}-40^{\circ})\tan( 90^{\circ}-10^{\circ})\right\}$

$=\frac{\sin 18^{\circ}}{\sin 18^{\circ}} +\tan 10^{\circ}\tan 40^{\circ}\cot 40^{\circ}\cot 10^{\circ}$

$=1+( 1)( 1)$

$=1+1$

$=2$

Hence, $\frac{\sin 18^{\circ}}{\cos 72^{\circ}}+\sqrt{3}\left\{\tan 10^{\circ} \tan 30^{\circ} \tan 40^{\circ} \tan 50^{\circ} \tan 80^{\circ}\right\}=2$. 

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Simply Easy Learning

Updated on: 10-Oct-2022

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