If $\cos \alpha=\frac{l \cos \beta+m}{l+m \cos \beta}$, then $\left(\frac{\tan \frac{\alpha}{2}}{\tan \frac{\beta}{2}}\right)^2=$

• Correct answer
  A

  $\frac{l-m}{l+m}$
• B

  $\frac{l+m}{l-m}$
• C

  $\frac{l^2-m^2}{l^2+m^2}$
• D

  $\sqrt{\frac{l-m}{l+m}}$

Hence, the correct answer is option: 'A'.