The minimum value of 27 tan^2 θ+3 cot^2 θ is

Correct answer: (B) 18. Given trigonometrical equation is 27 tan^2 θ+3 cot^2 θ ∴ (27 tan^2 θ+3 cot^2 θ)/2 ≥ √(27 tan^2 θ · 3 cot^2 θ) ⇒ (27 tan^2 θ+3 cot^2 θ)/2 ≥ √81 ⇒ 27 tan^2 θ+3 cot^2 θ ≥ 18 Hence, minimum value of given equation is 18

Q99 / 160 · TS EAMCET 2012 · Mathematics

The minimum value of $27 \tan ^2 \theta+3 \cot ^2 \theta$ is

A
$15$
Correct answer
B
$18$
C
$24$
D
$30$

Solution

Given trigonometrical equation is $27 \tan ^2 \theta+3 \cot ^2 \theta$ $\begin{aligned} & \therefore \quad \frac{27 \tan ^2 \theta+3 \cot ^2 \theta}{2} \geq \sqrt{27 \tan ^2 \theta \cdot 3 \cot ^2 \theta} \\ & \Rightarrow \frac{27 \tan ^2 \theta+3 \cot ^2 \theta}{2} \geq \sqrt{81} \\ & \Rightarrow 27 \tan ^2 \theta+3 \cot ^2 \theta \geq 18\end{aligned}$ Hence, minimum value of given equation is 18

Hence, the correct answer is option: 'B'.

The minimum value of $27 \tan ^2 \theta+3 \cot ^2 \theta$ is

Solution

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