The value of tan 20^{circ} + tan 40^{circ} + | vedclass

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(C) The given expression is $S = 	an 20^{\circ} + 	an 40^{\circ} + 	an 60^{\circ} + \cdots + 	an 180^{\circ}$.We know that $	an(180^{\circ} - 	h

Vedclass · Accepted Answer

$0$

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(C) The given expression is $S = 	an 20^{\circ} + 	an 40^{\circ} + 	an 60^{\circ} + \cdots + 	an 180^{\circ}$.We know that $	an(180^{\circ} - 	heta) = -	an 	heta$.Also,$	an 180^{\circ} = 0$.The series can be written as $	an 20^{\circ} + 	an 40^{\circ} + 	an 60^{\circ} + 	an 80^{\circ} + 	an 100^{\circ} + 	an 120^{\circ} + 	an 140^{\circ} + 	an 160^{\circ} + 	an 180^{\circ}$.Pairing the terms: $	an 160^{\circ} = 	an(180^{\circ} - 20^{\circ}) = -	an 20^{\circ}$.$	an 140^{\circ} = 	an(180^{\circ} - 40^{\circ}) = -	an 40^{\circ}$.$	an 120^{\circ} = 	an(180^{\circ} - 60^{\circ}) = -	an 60^{\circ}$.$	an 100^{\circ} = 	an(180^{\circ} - 80^{\circ}) = -	an 80^{\circ}$.Substituting these values: $S = (	an 20^{\circ} - 	an 20^{\circ}) + (	an 40^{\circ} - 	an 40^{\circ}) + (	an 60^{\circ} - 	an 60^{\circ}) + (	an 80^{\circ} - 	an 80^{\circ}) + 0$.$S = 0 + 0 + 0 + 0 + 0 = 0$.

The value of $\tan 20^{\circ} + \tan 40^{\circ} + \tan 60^{\circ} + \cdots + \tan 180^{\circ}$ is

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