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Evaluate:
$ \tan 7^{\circ} \tan 23^{\circ} \tan 60^{\circ} \tan 67^{\circ} \tan 83^{\circ} $
Given:
$\tan 7^{\circ} \tan 23^{\circ} \tan 60^{\circ} \tan 67^{\circ} \tan 83^{\circ}$.
To do:
We have to evaluate $\tan 7^{\circ} \tan 23^{\circ} \tan 60^{\circ} \tan 67^{\circ} \tan 83^{\circ}$.
Solution:
We know that,
$tan\ (90^{\circ}- \theta) = cot\ \theta$
$tan\ \theta \times \cot\ \theta=1$
Therefore,
$\tan 7^{\circ} \tan 23^{\circ} \tan 60^{\circ} \tan 67^{\circ} \tan 83^{\circ}=\tan 7^{\circ} \tan 23^{\circ} \tan 60^{\circ} \tan (90^{\circ}- 23^{\circ}) \tan (90^{\circ}-7^{\circ})$
$=\tan 7^{\circ} \tan 23^{\circ} (\sqrt3) \cot 23^{\circ} \cot 7^{\circ}$ (Since $\tan 60^{\circ}=\sqrt3$)
$=\sqrt3(\tan 7^{\circ} \cot 7^{\circ})(\tan 23^{\circ}\cot 23^{\circ})$
$=\sqrt3\times1\times1$
$=\sqrt3$
Hence, $\tan 7^{\circ} \tan 23^{\circ} \tan 60^{\circ} \tan 67^{\circ} \tan 83^{\circ}=\sqrt3$.