Evaluate the following:
$ \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} $

Given:

\( \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} \)

To do:

We have to evaluate \( \tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ} \).

Solution:  

We know that,

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

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

Therefore,

$\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=\tan (90^{\circ}-42^{\circ})\tan 23^{\circ}\tan 42^{\circ}\tan (90^{\circ}-23^{\circ})$

$=\tan 42^{\circ}\tan 23^{\circ}\cot 42^{\circ}\cot 23^{\circ}$

$=(\tan 42^{\circ}\cot 42^{\circ})(\tan 23^{\circ}\cot 23^{\circ})$

$=1\times1$

$=1$

Therefore, $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$.  

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Updated on: 10-Oct-2022

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