\[\underset{n\to \infty }{\mathop{\text{lim}\,}}\,\left[ \frac{1}{{{n}^{2}}}{{\sec }^{2}}\frac{1}{{{n}^{2}}}+\frac{2}{{{n}^{2}}}{{\sec }^{2}}\frac{4}{{{n}^{2}}}+.....+\frac{1}{n}{{\sec }^{2}}1 \right]\] equals [AIEEE 2005]

\[\underset{n\to \infty }{\mathop{\text{lim}\,}}\,\left[ \frac{1}{{{n}

1.	\[\underset{n\to \infty }{\mathop{\text{lim}\,}}\,\left[ \frac{1}{{{n}^{2}}}{{\sec }^{2}}\frac{1}{{{n}^{2}}}+\frac{2}{{{n}^{2}}}{{\sec }^{2}}\frac{4}{{{n}^{2}}}+.....+\frac{1}{n}{{\sec }^{2}}1 \right]\] equals [AIEEE 2005]
A.	\[\tan 1\]
B.	\[\frac{1}{2}\tan 1\]
C.	\[\frac{1}{2}\sec 1\]
D.	\[\frac{1}{2}\text{cosec}1\]
Answer» C. \[\frac{1}{2}\sec 1\]

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\[\underset{n\to \infty }{\mathop{\text{lim}\,}}\,\left[ \frac{1}{{{n}^{2}}}{{\sec }^{2}}\frac{1}{{{n}^{2}}}+\frac{2}{{{n}^{2}}}{{\sec }^{2}}\frac{4}{{{n}^{2}}}+.....+\frac{1}{n}{{\sec }^{2}}1 \right]\] equals [AIEEE 2005]

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