If frac{1+cos A}{1-cos A}=frac{m^{2}}{n^{2}}, | vedclass

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(B) Given $\frac{1+\cos A}{1-\cos A} = \frac{m^{2}}{n^{2}}$.Applying componendo and dividendo:$\frac{(1+\cos A) + (1-\cos A)}{(1+\cos A) - (1-\cos A)}

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$\pm \frac{2mn}{m^{2}-n^{2}}$

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(B) Given $\frac{1+\cos A}{1-\cos A} = \frac{m^{2}}{n^{2}}$.Applying componendo and dividendo:$\frac{(1+\cos A) + (1-\cos A)}{(1+\cos A) - (1-\cos A)} = \frac{m^{2}+n^{2}}{m^{2}-n^{2}}$$\frac{2}{2\cos A} = \frac{m^{2}+n^{2}}{m^{2}-n^{2}}$$\cos A = \frac{m^{2}-n^{2}}{m^{2}+n^{2}}$Since $\sin^{2} A = 1 - \cos^{2} A = 1 - \left(\frac{m^{2}-n^{2}}{m^{2}+n^{2}}ight)^{2} = \frac{(m^{2}+n^{2})^{2} - (m^{2}-n^{2})^{2}}{(m^{2}+n^{2})^{2}} = \frac{4m^{2}n^{2}}{(m^{2}+n^{2})^{2}}$Therefore,$\sin A = \pm \frac{2mn}{m^{2}+n^{2}}$.Finally,$	an A = \frac{\sin A}{\cos A} = \left(\pm \frac{2mn}{m^{2}+n^{2}}ight) / \left(\frac{m^{2}-n^{2}}{m^{2}+n^{2}}ight) = \pm \frac{2mn}{m^{2}-n^{2}}$.

If $\frac{1+\cos A}{1-\cos A}=\frac{m^{2}}{n^{2}}$,then $\tan A=$

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