If $\tan \left(\frac{\pi}{4}+\frac{\alpha}{2}\right)=\tan ^3\left(\frac{\pi}{4}+\frac{\beta}{2}\right)$, then $\frac{3+\sin ^2 \beta}{1+3 \sin ^2 \beta}=$

• A

  $\frac{\cos \beta}{\cos \alpha}$
• B

  $\frac{\cos ^3 \alpha}{\sin ^3 \beta}$
• Correct answer
  C

  $\frac{\sin \alpha}{\sin \beta}$
• D

  $\frac{\cos \alpha}{\cos \beta}$

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