$lim_{n \to \infty} [\frac{1}{n^{2}} \sec^{2} \frac{1}{n^{2}} + \frac{2}{n^{2}} \sec^{2} \frac{4}{n^{2}} \dots . . + \frac{1}{n} \sec^{2} 1]$ equals

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$\frac{1}{2} cosec 1$

$\frac{1}{2} \tan 1$

tan1

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

answer is D.

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Detailed Solution

$lim_{x \to \infty} [\frac{1}{n^{2}} \sec^{2} \frac{1}{n^{2}} + \frac{2}{n^{2}} \sec^{2} \frac{4}{n^{2}} + \frac{3}{n^{2}} \sec^{2} \frac{9}{n^{2}} + \dots + \frac{1}{n} \sec^{2} 1]$

$\Rightarrow lim_{x \to \infty} \frac{r}{n^{2}} \sec^{2} \frac{r^{2}}{n^{2}}$

$\Rightarrow \int_{0}^{1} x \sec^{2} x^{2} d x x^{2} = t \Rightarrow 2 x d x = d t d x = \frac{1}{2} d t \Rightarrow \frac{1}{2} \int_{0}^{1} \sec^{2} t d x = \frac{1}{2} (\tan t)_{0}^{1} = \frac{1}{2} \tan 1$