MCQ
$(\sec 2A + 1){\sec ^2}A = $
- A$\sec A$
- B$2\sec A$
- C$\sec 2A$
- ✓$2\sec 2A$
$= \left( {\frac{{1 + {{\tan }^2}A}}{{1 - {{\tan }^2}A}} + 1} \right)\,(1 + {\tan ^2}A)$
$ = \frac{{2\,(1 + {{\tan }^2}A)}}{{1 - {{\tan }^2}A}}$
$= 2\sec 2A.$
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$\begin{array}{*{20}{c}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1&3\end{array}$
$\begin{array}{*{20}{c}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,5&7&9&{11}\end{array}$
$\begin{array}{*{20}{c}}{13}&{15}&{17}&{19}&{21}&{23}\\.&.&.&.&.&.\\.&.&.&.&.&.\\.&.&.&.&.&.\end{array}$
Then the sum of ${n^{th}}$ row is