If $\operatorname{Tan}^{-1} \frac{1}{3}+\operatorname{Tan}^{-1} \frac{1}{7}+\operatorname{Tan}^{-1} \frac{1}{13}+\ldots+\operatorname{Tan}^{-1} \frac{1}{n^2+n+1}=\operatorname{Tan}^{-1} \theta$, then $\theta=$

• Correct answer
  A

  $\frac{n}{n+2}$
• B

  $\frac{n}{n+1}$
• C

  $\frac{n+1}{n+2}$
• D

  $\frac{n-1}{n+2}$

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