The value of tan {20^circ} + 2tan {50^circ} - | vedclass

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(B) We have the expression $	an {20^\circ} + 2	an {50^\circ} - 	an {70^\circ}$.Rewrite the expression as $(	an {20^\circ} - 	an {70^\circ}) + 2

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(B) We have the expression $	an {20^\circ} + 2	an {50^\circ} - 	an {70^\circ}$.Rewrite the expression as $(	an {20^\circ} - 	an {70^\circ}) + 2	an {50^\circ}$.Using $	an 	heta = \frac{\sin 	heta}{\cos 	heta}$,we get $\frac{\sin {20^\circ}}{\cos {20^\circ}} - \frac{\sin {70^\circ}}{\cos {70^\circ}} + 2	an {50^\circ}$.$= \frac{\sin {20^\circ} \cos {70^\circ} - \cos {20^\circ} \sin {70^\circ}}{\cos {20^\circ} \cos {70^\circ}} + 2	an {50^\circ}$.Using $\sin(A-B) = \sin A \cos B - \cos A \sin B$ and $2\cos A \cos B = \cos(A+B) + \cos(A-B)$:$= \frac{\sin(20^\circ - 70^\circ)}{\frac{1}{2}[\cos(70^\circ + 20^\circ) + \cos(70^\circ - 20^\circ)]} + 2	an {50^\circ}$.$= \frac{2\sin(-50^\circ)}{\cos {90^\circ} + \cos {50^\circ}} + 2	an {50^\circ}$.Since $\cos {90^\circ} = 0$,this becomes $\frac{-2\sin {50^\circ}}{\cos {50^\circ}} + 2	an {50^\circ}$.$= -2	an {50^\circ} + 2	an {50^\circ} = 0$.

The value of $\tan {20^\circ} + 2\tan {50^\circ} - \tan {70^\circ}$ is equal to

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