tan 20^circ tan 40^circ tan 60^circ tan 80^ci | vedclass

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Question

(C) We use the identity $	an 	heta 	an(60^\circ - 	heta) 	an(60^\circ + 	heta) = 	an 3	heta$.For $	heta = 20^\circ$,we have $	an 20^\circ

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) We use the identity $	an 	heta 	an(60^\circ - 	heta) 	an(60^\circ + 	heta) = 	an 3	heta$.For $	heta = 20^\circ$,we have $	an 20^\circ 	an(60^\circ - 20^\circ) 	an(60^\circ + 20^\circ) = 	an(3 	imes 20^\circ) = 	an 60^\circ$.Thus,$	an 20^\circ 	an 40^\circ 	an 80^\circ = 	an 60^\circ = \sqrt{3}$.Now,the given expression is $	an 20^\circ 	an 40^\circ 	an 60^\circ 	an 80^\circ$.Substituting the value of the product $	an 20^\circ 	an 40^\circ 	an 80^\circ$,we get:$(	an 20^\circ 	an 40^\circ 	an 80^\circ) 	imes 	an 60^\circ = \sqrt{3} 	imes \sqrt{3} = 3$.

$\tan 20^\circ \tan 40^\circ \tan 60^\circ \tan 80^\circ = $

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