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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Similar Questions

$\max_{0 \le x \le \pi} (16 \sin^2(\frac{x}{2}) \cos^3(\frac{x}{2}))$ is equal to:

The minimum value of $8 \cos^2 x + 18 \sec^2 x$ for all $x \in R$ wherever it is defined,is:

If $x + y + z = 180^\circ$,then $\cos 2x + \cos 2y - \cos 2z$ is equal to

If $A+B+C=270^{\circ}$,then $\cos 2A+\cos 2B+\cos 2C$ is equal to:

If $A+B+C=\frac{3 \pi}{2}$,then $4 \sin A \sin B \sin C+\cos 2 A+\cos 2 B+\cos 2 C=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module