Question

The minimum value of $9{\tan }^2\theta +4{\cot }^2\theta$ is equal to

• A. 13
• B. 9
• C. 6
• D. 12

Answer 1

Answer: (D)

Since A.M. $\ge$ G.M., therefore
$\frac{9{\tan }^2\theta +4{\cot }^2\theta }{2}\ge \sqrt{9{\tan }^2\theta \times 4{\cot }^2\theta }$
$\Rightarrow 9{\tan }^2\theta +4{\cot }^2\theta \ge 2\times \sqrt{36}$
$=12.$
Hence the required minimum value is 12.
Alternatively
$9{\tan }^2\theta +4{\cot }^2\theta ={\left(3\tan \theta -2\cot \theta \right)}^2+12\tan \theta \cot \theta \ge 12.$