The least value of E = frac{25sec^4 x - 50sec | vedclass

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(B) Given $E = \frac{25\sec^4 x - 50\sec^2 x + 74}{	an^2 x}$.Using the identity $\sec^2 x = 1 + 	an^2 x$,we can rewrite the numerator:$E = \frac{25(

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(B) Given $E = \frac{25\sec^4 x - 50\sec^2 x + 74}{	an^2 x}$.Using the identity $\sec^2 x = 1 + 	an^2 x$,we can rewrite the numerator:$E = \frac{25(\sec^2 x - 1)^2 + 49}{	an^2 x}$Since $\sec^2 x - 1 = 	an^2 x$,we have:$E = \frac{25(	an^2 x)^2 + 49}{	an^2 x}$$E = 25	an^2 x + \frac{49}{	an^2 x}$Using the Arithmetic Mean-Geometric Mean inequality $(AM \ge GM)$:$E \ge 2 \sqrt{25	an^2 x \cdot \frac{49}{	an^2 x}}$$E \ge 2 \sqrt{25 \cdot 49}$$E \ge 2 \cdot 5 \cdot 7 = 70$Thus,the minimum value is $70$.

The least value of $E = \frac{25\sec^4 x - 50\sec^2 x + 74}{\tan^2 x}$ is

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