(D) Given expression: $(\sec 2A + 1)\sec^2 A$Substitute $\sec 2A = \frac{1}{\cos 2A} = \frac{1+\tan^2 A}{1-\tan^2 A}$ and $\sec^2 A = 1+\tan^2 A$:$= \

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(D) Given expression: $(\sec 2A + 1)\sec^2 A$Substitute $\sec 2A = \frac{1}{\cos 2A} = \frac{1+\tan^2 A}{1-\tan^2 A}$ and $\sec^2 A = 1+\tan^2 A$:$= \left( \frac{1+\tan^2 A}{1-\tan^2 A} + 1 \right)(1+\tan^2 A)$$= \left( \frac{1+\tan^2 A + 1 - \tan^2 A}{1-\tan^2 A} \right)(1+\tan^2 A)$$= \left( \frac{2}{1-\tan^2 A} \right)(1+\tan^2 A)$$= \frac{2(1+\tan^2 A)}{1-\tan^2 A} = 2 \sec 2A$

Class 11 Mathematics Trigonometrical Ratios, Functions and Identities Trigonometrical ratios of multiple and sub-multiple angles

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

A
$\sec A$
B
$2\sec A$
C
$\sec 2A$
D
$2\sec 2A$

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Trigonometrical Ratios, Functions and Identities

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Trigonometrical ratios of multiple and sub-multiple angles

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Find the value of $\tan \frac{\pi}{8}$.

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

Similar Questions

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