Question

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

• 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}}} $

Answer 1

Answer: (B)

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}}} $