Prove that:$ \tan 20^{\circ} \tan 35^{\circ} \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ}=1 $

To do:

We have to prove that $\tan 20^{\circ} \tan 35^{\circ} \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ}=1$.

Solution:  

We know that,

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

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

Therefore,

$\tan 20^{\circ} \tan 35^{\circ} \tan 45^{\circ} \tan 55^{\circ} \tan 70^{\circ}=\tan 20^{\circ} \tan 35^{\circ} \tan 45^{\circ} \tan (90^{\circ}- 35^{\circ}) \tan (90^{\circ}-20^{\circ})$

$=\tan 20^{\circ} \tan 35^{\circ} (1) \cot 35^{\circ} \cot 20^{\circ}$     (Since $\tan 45^{\circ}=1$)

$=(\tan 20^{\circ} \cot 20^{\circ})(\tan 35^{\circ}\cot 35^{\circ})$

$=1\times1$

$=1$

Hence proved.

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Updated on: 10-Oct-2022

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