Question

$\underset{n \rightarrow \infty}{lim} \bigg [\frac{1}{n^2}sec^2\frac{1}{n^2}+\frac{2}{n^2}sec^2\frac{4}{n^2}..........+\frac{1}{n}sec^21\bigg]$ equals
(a) $\frac{1}{2}$sec1
(b) $\frac{1}{2}$cosec 1
(c) tan 1
(d) $\frac{1}{2}$tan 1

Answer 1

I don't know the answer sir