$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots+\frac{1}{n^2} \sec ^2 1\right]=$

• A

  $\operatorname{Tan}^{-1} 1$
• B

  $\frac{1}{2} \operatorname{Tan}^{-1} 1$
• Correct answer
  C

  $\frac{1}{2} \tan 1$
• D

  $\frac{1}{2} \sec 1$

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