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

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(C) We know that $\sqrt{3} = 	an 60^{\circ} = 	an(40^{\circ} + 20^{\circ})$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an

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$\sqrt{3}$

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(C) We know that $\sqrt{3} = 	an 60^{\circ} = 	an(40^{\circ} + 20^{\circ})$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,we have:$\sqrt{3} = \frac{	an 40^{\circ} + 	an 20^{\circ}}{1 - 	an 40^{\circ} 	an 20^{\circ}}$.Cross-multiplying gives:$\sqrt{3}(1 - 	an 40^{\circ} 	an 20^{\circ}) = 	an 40^{\circ} + 	an 20^{\circ}$.$\sqrt{3} - \sqrt{3} 	an 40^{\circ} 	an 20^{\circ} = 	an 40^{\circ} + 	an 20^{\circ}$.Rearranging the terms:$	an 20^{\circ} + 	an 40^{\circ} + \sqrt{3} 	an 20^{\circ} 	an 40^{\circ} = \sqrt{3}$.

$\tan\, 20^{\circ} + \tan\, 40^{\circ} + \sqrt{3}\, \tan\, 20^{\circ} \tan\, 40^{\circ}$ is equal to

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