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
$\frac{9 \tan^{2} \theta + 4 \cot^{2} \theta}{2} \ge \sqrt{9 \tan^{2} \theta \times4 \cot^{2} \theta}$
$ \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. $