If m tan (theta - 30^circ) = n tan (theta + 1 | vedclass

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Question

(A) Given: $m 	an (	heta - 30^\circ) = n 	an (	heta + 120^\circ)$.Rearranging the terms,we get: $\frac{m}{n} = \frac{	an (	heta + 120^\circ)}{

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$2 \cos 2	heta$

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(A) Given: $m 	an (	heta - 30^\circ) = n 	an (	heta + 120^\circ)$.Rearranging the terms,we get: $\frac{m}{n} = \frac{	an (	heta + 120^\circ)}{	an (	heta - 30^\circ)}$.Applying the Componendo and Dividendo rule:$\frac{m + n}{m - n} = \frac{	an (	heta + 120^\circ) + 	an (	heta - 30^\circ)}{	an (	heta + 120^\circ) - 	an (	heta - 30^\circ)}$.Using the identity $	an A + 	an B = \frac{\sin(A+B)}{\cos A \cos B}$ and $	an A - 	an B = \frac{\sin(A-B)}{\cos A \cos B}$,we get:$\frac{m + n}{m - n} = \frac{\sin((	heta + 120^\circ) + (	heta - 30^\circ))}{\sin((	heta + 120^\circ) - (	heta - 30^\circ))}$.$\frac{m + n}{m - n} = \frac{\sin(2	heta + 90^\circ)}{\sin(150^\circ)}$.Since $\sin(2	heta + 90^\circ) = \cos 2	heta$ and $\sin(150^\circ) = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac{1}{2}$.Therefore,$\frac{m + n}{m - n} = \frac{\cos 2	heta}{1/2} = 2 \cos 2	heta$.

If $m \tan (\theta - 30^\circ) = n \tan (\theta + 120^\circ)$,then $\frac{m + n}{m - n} = $

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