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)

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

Answer 2

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, $\tan A=\frac{2\tan (A/2)}{1-{{\tan }^{2}}(A/2)}$
$=\pm \frac{2(n/m)}{1-({{n}^{2}}/{{m}^{2}})}$
$=\pm \frac{2nm}{{{m}^{2}}-{{n}^{2}}}$