Question

If $\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$, then the value of $\tan A=$

• A. $\frac{x^2+y^2}{x^2-y^2}$
• B. $\frac{2xy}{x^2-y^2}$
• C. $\frac{2xy}{x^2+y^2}$
• D. $\frac{2xy}{y^2-x^2}$

Answer 1

The correct option is B $\frac{2xy}{x^2-y^2}$

$1+\cos A=2{\cos }^2\left(A/2\right)$

$1-\cos A=2{\sin }^2\left(A/2\right)$

So $\sqrt{\frac{1+\cos A}{1-\cos A}}=\sqrt{\frac{2{\cos }^2\frac{A}{2}}{2{\sin }^2\frac{A}{2}}}=cot\frac{A}{2}=\frac{x}{y}$

Now, $\tan A=\frac{2\tan \frac{A}{2}}{1-{\tan }^2\frac{A}{2}}$

$\therefore \tan A=\frac{2\tan \frac{A}{2}}{1-{\tan }^2\frac{A}{2}}=\frac{2\times \left(\frac{1}{\cot \frac{A}{2}}\right)}{1-{\left(\frac{1}{\cot \frac{A}{2}}\right)}^2}=\frac{2\cot \frac{A}{2}}{{\cot }^2\frac{A}{2}-1}=\frac{2\frac{x}{y}}{{\left(\frac{x}{y}\right)}^2-1}=\frac{2xy}{x^2-y^2}$