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

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

Vedclass · Accepted Answer

$2\sec 2A$

Vedclass · Answer

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

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

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