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

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(B) We know that the formula for $	an(A + B)$ is: $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ Substituting $A = 20^\circ$ and $B = 40^\

Vedclass · Accepted Answer

$\sqrt{3}$

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(B) We know that the formula for $	an(A + B)$ is: $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ Substituting $A = 20^\circ$ and $B = 40^\circ$: $	an(20^\circ + 40^\circ) = \frac{	an 20^\circ + 	an 40^\circ}{1 - 	an 20^\circ 	an 40^\circ}$ Since $	an(60^\circ) = \sqrt{3}$,we have: $\sqrt{3} = \frac{	an 20^\circ + 	an 40^\circ}{1 - 	an 20^\circ 	an 40^\circ}$ Multiplying both sides by $(1 - 	an 20^\circ 	an 40^\circ)$: $\sqrt{3} - \sqrt{3} 	an 20^\circ 	an 40^\circ = 	an 20^\circ + 	an 40^\circ$ Rearranging the terms: $	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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