MCQ
$9{\tan ^2}\theta + 4{\cot ^2}\theta $ ની ન્યૂનતમ કિમત મેળવો.
- 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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