Evaluate:
$ \tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ} $


Given:

$\tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ}$

To do:

We have to evaluate $\tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ}$.

Solution:  

We know that,

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

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

Therefore,

$\tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ}=\tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan (90^{\circ}- 40^{\circ}) \tan (90^{\circ}-35^{\circ})$

$=\tan 35^{\circ} \tan 40^{\circ} (1) \cot 40^{\circ} \cot 35^{\circ}$     (Since $\tan 45^{\circ}=1$)

$=(\tan 35^{\circ} \cot 35^{\circ})(\tan 40^{\circ}\cot 40^{\circ})$

$=1\times1$

$=1$ 

Hence, $\tan 35^{\circ} \tan 40^{\circ} \tan 45^{\circ} \tan 50^{\circ} \tan 55^{\circ}=1$.

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Updated on: 10-Oct-2022

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