Question

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

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