A) \[\pm \frac{2mn}{{{m}^{2}}+{{n}^{2}}}\]
B) \[\pm \frac{2mn}{{{m}^{2}}-{{n}^{2}}}\]
C) \[\frac{{{m}^{2}}+{{n}^{2}}}{{{m}^{2}}-{{n}^{2}}}\]
D) \[\frac{{{m}^{2}}-{{n}^{2}}}{{{m}^{2}}+{{n}^{2}}}\]
Correct Answer: B
Solution :
Given, \[\frac{1+\cos A}{1-\cos A}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[\frac{{{\cos }^{2}}A/2}{{{\sin }^{2}}A/2}=\frac{{{m}^{2}}}{{{n}^{2}}}\] \[\Rightarrow \] \[{{\tan }^{2}}\frac{A}{2}=\frac{{{n}^{2}}}{{{m}^{2}}}\] \[\Rightarrow \] \[\tan \frac{A}{2}=\pm \frac{n}{m}\] Now, \[\tan A=\frac{2\tan (A/2)}{1-{{\tan }^{2}}(A/2)}\] \[=\pm \frac{2(n/m)}{1-({{n}^{2}}/{{m}^{2}})}\] \[=\pm \frac{2nm}{{{m}^{2}}-{{n}^{2}}}\]
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