If m cdot tan (theta-30^{circ})=n cdot tan (t | vedclass

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(A) Given: $m 	an (	heta-30^{\circ})=n 	an (	heta+120^{\circ})$Since $	an (	heta+120^{\circ}) = 	an (	heta+30^{\circ}+90^{\circ}) = -\cot (	h

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$2 \cos 2 	heta$

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(A) Given: $m 	an (	heta-30^{\circ})=n 	an (	heta+120^{\circ})$Since $	an (	heta+120^{\circ}) = 	an (	heta+30^{\circ}+90^{\circ}) = -\cot (	heta+30^{\circ})$,$\frac{n}{m} = \frac{	an (	heta-30^{\circ})}{-\cot (	heta+30^{\circ})} = -	an (	heta-30^{\circ}) 	an (	heta+30^{\circ})$Using $	an (A-B) 	an (A+B) = \frac{	an^2 A - 	an^2 B}{1 - 	an^2 A 	an^2 B}$,we get:$\frac{n}{m} = -\left( \frac{	an^2 	heta - 	an^2 30^{\circ}}{1 - 	an^2 	heta 	an^2 30^{\circ}} ight) = \frac{	an^2 30^{\circ} - 	an^2 	heta}{1 - 	an^2 	heta 	an^2 30^{\circ}}$Now,$\frac{m+n}{m-n} = \frac{1 + n/m}{1 - n/m} = \frac{1 + \frac{	an^2 30^{\circ} - 	an^2 	heta}{1 - 	an^2 	heta 	an^2 30^{\circ}}}{1 - \frac{	an^2 30^{\circ} - 	an^2 	heta}{1 - 	an^2 	heta 	an^2 30^{\circ}}}$$= \frac{1 - 	an^2 	heta 	an^2 30^{\circ} + 	an^2 30^{\circ} - 	an^2 	heta}{1 - 	an^2 	heta 	an^2 30^{\circ} - 	an^2 30^{\circ} + 	an^2 	heta} = \frac{(1 - 	an^2 	heta)(1 + 	an^2 30^{\circ})}{(1 + 	an^2 	heta)(1 - 	an^2 30^{\circ})}$$= \frac{\cos 2	heta}{\cos 2(30^{\circ})} = \frac{\cos 2	heta}{1/2} = 2 \cos 2	heta$

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

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