The minimum value of 5 tan^2 alpha + frac{9}{ | vedclass

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

Question

(B) Given expression: $5 	an^2 \alpha + \frac{9}{	an^2 \alpha} + 4 \sec^2 \alpha$ Using the identity $\sec^2 \alpha = 1 + 	an^2 \alpha$,we get: $=

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(B) Given expression: $5 	an^2 \alpha + \frac{9}{	an^2 \alpha} + 4 \sec^2 \alpha$ Using the identity $\sec^2 \alpha = 1 + 	an^2 \alpha$,we get: $= 5 	an^2 \alpha + 9 \cot^2 \alpha + 4(1 + 	an^2 \alpha)$ $= 9 	an^2 \alpha + 9 \cot^2 \alpha + 4$ Since $AM \geq GM$ for positive real numbers,we have: $\frac{9 	an^2 \alpha + 9 \cot^2 \alpha}{2} \geq \sqrt{9 	an^2 \alpha \cdot 9 \cot^2 \alpha}$ $9 	an^2 \alpha + 9 \cot^2 \alpha \geq 2 \cdot \sqrt{81 \cdot 1}$ $9 	an^2 \alpha + 9 \cot^2 \alpha \geq 18$ Adding $4$ to both sides: $9 	an^2 \alpha + 9 \cot^2 \alpha + 4 \geq 18 + 4 = 22$ Therefore,the minimum value is $22$.

The minimum value of $5 \tan^2 \alpha + \frac{9}{\tan^2 \alpha} + 4 \sec^2 \alpha$ is:

Similar Questions

If $A + B + C = \pi$ and $\sin \left( A + \frac{C}{2} \right) = k \sin \frac{C}{2}$,then $\tan \frac{A}{2} \tan \frac{B}{2} = $

If the equation $\cos^4 \theta + \sin^4 \theta + \lambda = 0$ has real solutions for $\theta$,then $\lambda$ lies in the interval

If $A+B+C=4S$,then $\cos (2S-A)+\cos (2S-B)-\cos (2S-C)-\cos 2S=$

For $A, B$ and $C$,if $A+B+C=0$,then $\sin(2A) + \sin(2B) + \sin(2C)$ is equal to

If $A = \sin^2 x + \cos^4 x$,then for all real $x :$

Vedclass Test Series

Exam Paper Generator

Online Exam Module