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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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