(sec 2A + 1){sec ^2}A = | vedclass

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(D) Given expression: $(\sec 2A + 1){\sec ^2}A$We know that $\sec 2A = \frac{1}{\cos 2A} = \frac{1 + 	an^2 A}{1 - 	an^2 A}$ and $\sec^2 A = 1 + 	an

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$2\sec 2A$

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(D) Given expression: $(\sec 2A + 1){\sec ^2}A$We know that $\sec 2A = \frac{1}{\cos 2A} = \frac{1 + 	an^2 A}{1 - 	an^2 A}$ and $\sec^2 A = 1 + 	an^2 A$.Substituting these into the expression:$= \left( \frac{1 + 	an^2 A}{1 - 	an^2 A} + 1 ight) (1 + 	an^2 A)$$= \left( \frac{1 + 	an^2 A + 1 - 	an^2 A}{1 - 	an^2 A} ight) (1 + 	an^2 A)$$= \left( \frac{2}{1 - 	an^2 A} ight) (1 + 	an^2 A)$$= 2 \cdot \frac{1 + 	an^2 A}{1 - 	an^2 A}$Since $\cos 2A = \frac{1 - 	an^2 A}{1 + 	an^2 A}$,it follows that $\sec 2A = \frac{1 + 	an^2 A}{1 - 	an^2 A}$.Therefore,the expression becomes $2 \sec 2A$.

$(\sec 2A + 1){\sec ^2}A = $

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