The minimum value of 9tan^2theta + 4cot^2thet | vedclass

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

Question

(D) We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for two positive numbers $a$ and $b$,which states that $\frac{a+b}{2} \ge \sqrt

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(D) We use the Arithmetic Mean-Geometric Mean $(AM \ge GM)$ inequality for two positive numbers $a$ and $b$,which states that $\frac{a+b}{2} \ge \sqrt{ab}$.Let $a = 9	an^2	heta$ and $b = 4\cot^2	heta$.Then,$\frac{9	an^2	heta + 4\cot^2	heta}{2} \ge \sqrt{9	an^2	heta \cdot 4\cot^2	heta}$.Since $	an	heta \cdot \cot	heta = 1$,we have $\sqrt{9 \cdot 4 \cdot 1} = \sqrt{36} = 6$.Therefore,$\frac{9	an^2	heta + 4\cot^2	heta}{2} \ge 6$.Multiplying both sides by $2$,we get $9	an^2	heta + 4\cot^2	heta \ge 12$.Thus,the minimum value is $12$.

The minimum value of $9\tan^2\theta + 4\cot^2\theta$ is

Similar Questions

If $\sin \theta + \sin 2\theta + \sin 3\theta = \sin \alpha$ and $\cos \theta + \cos 2\theta + \cos 3\theta = \cos \alpha$,then $\theta$ is equal to

Two towers $A$ and $B$ have heights $45 \, m$ and $15 \, m$ respectively. The angle of elevation from the bottom of tower $B$ to the top of tower $A$ is $60^{\circ}$. If the angle of elevation from the bottom of tower $A$ to the top of tower $B$ is $\theta$,then the value of $\sin \theta$ is:

Which of the following is correct?

The maximum value of $(7 \cos\theta + 24 \sin\theta) \times (7 \sin\theta - 24 \cos\theta)$ for every $\theta \in R$.

If $\sin \theta = \frac{12}{13}$ where $0 < \theta < \frac{\pi}{2}$ and $\cos \phi = -\frac{3}{5}$ where $\pi < \phi < \frac{3\pi}{2}$,then $\sin(\theta + \phi)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module