Question

Minimum value of 27 tan^2 theta+ 3 cot^2 theta

Answer 1

Answer:

Step-by-step explanation:

You can use the concept of AM >= GM to solve this

take numbers as tan^2theta,tan^2theta,tan^2theta….27 times and cot^2theta,cot^2theta,cot^2theta

So we will get

[27 tan^2 theta + 3 cot^2 theta]/ 30 >= {27tan^2 theta* 3 cot^2 theta}1/2

27 tan^2 theta + 3 cot^2 theta >= 9 * 30

Answer 2

Given:

$27 \tan^2 \theta$ + $3\cot^2 \theta$

We have to find, the minimum value of $27 \tan^2 \theta$ + $3\cot^2 \theta$ is:

Solution:

We know that,

AM ≥ GM

∴ $\dfrac{27 \tan^2 \theta+3\cot^2 \theta}{2}$ ≥ $\sqrt{27 \tan^2 \theta.3\cot^2 \theta}$ [∴ AM = $\dfrac{a+b}{2}$ and GM = $\sqrt{ab}$]

Using the trigonometric identity:

$\cot A$ = $\dfrac{1}{ \tan A}$

⇒ $\dfrac{27 \tan^2 \theta+3\cot^2 \theta}{2}$ ≥ $\sqrt{81 \tan^2 \theta.\dfrac{1}{ \tan^2 \theta} }$

⇒ $\dfrac{27 \tan^2 \theta+3\cot^2 \theta}{2}$ ≥ $\sqrt{81 }$

⇒ $\dfrac{27 \tan^2 \theta+3\cot^2 \theta}{2}$ ≥ 9

Multiplying both sides by 2, we get

$27 \tan^2 \theta$ + $3\cot^2 \theta$ ≥ 9 × 2

⇒ $27 \tan^2 \theta$ + $3\cot^2 \theta$ ≥ 18

∴ The minimum value of $27 \tan^2 \theta$ + $3\cot^2 \theta$ = 18

Thus, the minimum value of "$27 \tan^2 \theta$ + $3\cot^2 \theta$ is equal to 18".