MCQ
The minimum value of $9{\tan ^2}\theta + 4{\cot ^2}\theta $ is
- A$13$
- B$9$
- C$6$
- ✓$12$
$ \Rightarrow \frac{{9{{\tan }^2}\theta + 4{{\cot }^2}\theta }}{2} \ge \sqrt {4{{\cot }^2}\theta .9{{\tan }^2}\theta } $
$ \Rightarrow 9{\tan ^2}\theta + 4{\cot ^2}\theta \ge 12$
Therefore, the minimum value is $12.$
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$(A)$ $e_1^2+e_2^2=\frac{43}{40}$
$(B)$ $e_1 e_2=\frac{\sqrt{7}}{2 \sqrt{10}}$
$(C)$ $\left|e_1^2-e_2^2\right|=\frac{5}{8}$
$(D)$ $e_1 e_2=\frac{\sqrt{3}}{4}$