Q118 / 180 · COMEDK 2019 · Mathematics
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$\frac{2 x y}{x^{2}+y^{2}}$
- Correct answerC$\frac{2 x y}{x^{2}-y^{2}}$
- D$\frac{2 x y}{y^{2}-x^{2}}$
Solution
We have,
$\sqrt{\frac{1+\cos A}{1-\cos A}}=\frac{x}{y}$
$\Rightarrow \sqrt{\frac{2 \cos ^{2} \frac{A}{2}}{2 \sin ^{2} \frac{A}{2}}}=\frac{x}{y}$
$\Rightarrow \quad \cot \frac{A}{2}=\frac{x}{y} \Rightarrow \tan \frac{A}{2}=\frac{y}{x}$
Now, $\tan A=\frac{2 \tan \frac{A}{2}}{1-\tan ^{2} \frac{A}{2}}=\frac{2\left(\frac{y}{x}\right)}{1-\left(\frac{y}{x}\right)^{2}}=\frac{2 x y}{x^{2}-y^{2}}$
Hence, the correct answer is option: 'C'.