$tan 60^{\circ} = \frac{1}{\sqrt{3}}$

True

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False

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Solution

The correct option is B False

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Similar questions

tanA+tan(60+A)+tan(60-A)=3tan3A

[\frac{tan 60^{\circ} + 1}{tan 60^{\circ} - 1}] = \frac{1 + tan 30^{\circ}}{1 - tan 30^{\circ}}

°

\frac{\tan 60 ° - \tan 30 °}{1 + \tan 60 ° \tan 30 °}

\frac{tan 60^{\circ} - tan 30^{\circ}}{1 + tan 60^{\circ} tan 30^{\circ}}

Show that:

(i) $\frac{1 - sin 60^{\circ}}{cos 60^{\circ}} = \frac{tan 60^{\circ} - 1}{tan 60^{\circ} + 1}$

(ii) $\frac{cos 30^{\circ} + sin 60^{\circ}}{1 + sin 30^{\circ} + cos 60^{\circ}} = cos 30^{\circ}$

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