(D) આપેલ પદાવલિ: $(\sec 2A + 1)\sec^2 A$$\sec 2A = \frac{1}{\cos 2A} = \frac{1+\tan^2 A}{1-\tan^2 A}$ અને $\sec^2 A = 1+\tan^2 A$ મૂકતા:$= \left( \fra

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(D) આપેલ પદાવલિ: $(\sec 2A + 1)\sec^2 A$$\sec 2A = \frac{1}{\cos 2A} = \frac{1+\tan^2 A}{1-\tan^2 A}$ અને $\sec^2 A = 1+\tan^2 A$ મૂકતા:$= \left( \frac{1+\tan^2 A}{1-\tan^2 A} + 1 \right)(1+\tan^2 A)$$= \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$