If $\operatorname{Tan}^{-1}\left[\frac{1}{1+2}\right]+\operatorname{Tan}^{-1}\left[\frac{1}{1+(2)(3)}\right]+\operatorname{Tan}^{-1}\left[\frac{1}{1+(3)(4)}\right]+\cdots \cdots+\operatorname{Tan}^{-1}\left[\frac{1}{1+n(n+1)}\right]=\operatorname{Tan}^{-1} \theta$, then $\theta=$

• A

  $\frac{n}{n+1}$
• B

  $\frac{n+1}{n+2}$
• C

  $\frac{n+2}{n+1}$
• Correct answer
  D

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

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