tan 20^circ + tan 40^circ + sqrt{3} tan 20^ci | vedclass

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(B) We know the trigonometric identity for the tangent of a sum: $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Let $A = 20^\circ$ and $B =

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$\sqrt{3}$

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(B) We know the trigonometric identity for the tangent of a sum: $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Let $A = 20^\circ$ and $B = 40^\circ$.Then,$	an(20^\circ + 40^\circ) = 	an 60^\circ = \sqrt{3}$.Substituting these into the identity:$\sqrt{3} = \frac{	an 20^\circ + 	an 40^\circ}{1 - 	an 20^\circ 	an 40^\circ}$.Cross-multiplying gives:$\sqrt{3}(1 - 	an 20^\circ 	an 40^\circ) = 	an 20^\circ + 	an 40^\circ$.$\sqrt{3} - \sqrt{3} 	an 20^\circ 	an 40^\circ = 	an 20^\circ + 	an 40^\circ$.Rearranging the terms,we get:$	an 20^\circ + 	an 40^\circ + \sqrt{3} 	an 20^\circ 	an 40^\circ = \sqrt{3}$.

$\tan 20^\circ + \tan 40^\circ + \sqrt{3} \tan 20^\circ \tan 40^\circ = $

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