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