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

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(B) Given expression: $	an 20^\circ + 2	an 50^\circ - 	an 70^\circ$$= (	an 20^\circ - 	an 70^\circ) + 2	an 50^\circ$$= (\frac{\sin 20^\circ}{\co

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(B) Given expression: $	an 20^\circ + 2	an 50^\circ - 	an 70^\circ$$= (	an 20^\circ - 	an 70^\circ) + 2	an 50^\circ$$= (\frac{\sin 20^\circ}{\cos 20^\circ} - \frac{\sin 70^\circ}{\cos 70^\circ}) + 2	an 50^\circ$$= \frac{\sin 20^\circ \cos 70^\circ - \cos 20^\circ \sin 70^\circ}{\cos 20^\circ \cos 70^\circ} + 2	an 50^\circ$Using $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$= \frac{\sin(20^\circ - 70^\circ)}{\cos 20^\circ \cos 70^\circ} + 2	an 50^\circ$$= \frac{\sin(-50^\circ)}{\cos 20^\circ \sin 20^\circ} + 2	an 50^\circ$ (since $\cos 70^\circ = \sin 20^\circ$)$= \frac{-2\sin 50^\circ}{2\sin 20^\circ \cos 20^\circ} + 2	an 50^\circ$$= \frac{-2\sin 50^\circ}{\sin 40^\circ} + 2	an 50^\circ$Since $\sin 40^\circ = \cos 50^\circ$:$= \frac{-2\sin 50^\circ}{\cos 50^\circ} + 2	an 50^\circ$$= -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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