tan,, 20^o + tan,, 40^o + sqrt 3,, | vedclass

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Question

$ \sqrt{3} =	an 60^{\circ}=	an \left(40^{\circ}+20^{\circ}ight) $

$=\frac{	an 40^{\circ}+	an 20^{\circ}}{1-	an 40^{\circ} 	an 20^{\circ}} $

Vedclass · Accepted Answer

$\sqrt 3$

Vedclass · Answer

$ \sqrt{3} =	an 60^{\circ}=	an \left(40^{\circ}+20^{\circ}ight) $

$=\frac{	an 40^{\circ}+	an 20^{\circ}}{1-	an 40^{\circ} 	an 20^{\circ}} $

$ 	herefore  \sqrt{3}-\sqrt{3} 	an 40^{\circ} 	an 20^{\circ}=	an 40^{\circ}+	an 20^{\circ} $

Hence $	an {40^\circ } + 	an {20^\circ } + \sqrt 3 	an {40^\circ }	an {20^\circ }$

$=\sqrt{3}$

$tan\,\, 20^o + tan\,\, 40^o + \sqrt 3\,\, tan\,\, 20^o tan\,\, 40^o$ is equal to

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