Question

The least value of $\frac{1}{2}+\frac{2}{3}{\text{cosec}}^2\theta +\frac{3}{8}{\sec }^2\theta$ is

• A. $\frac{13}{24}$
• B. $\frac{61}{48}$
• C. $\frac{61}{25}$
• D. $\frac{61}{24}$

Answer 1

The correct option is D $\frac{61}{24}$

$\frac{1}{2}+\frac{2}{3}{\text{cosec}}^2\theta +\frac{3}{8}{\sec }^2\theta$

$=\frac{1}{2}+\frac{2}{3}+\frac{2}{3}{\cot }^2\theta +\frac{3}{8}+\frac{3}{8}{\tan }^2\theta$

$=\frac{37}{24}+\frac{2}{3}{\cot }^2\theta +\frac{3}{8}{\tan }^2\theta$

Now, AM$\ge$GM
$\frac{\frac{2}{3}{\cot }^2\theta +\frac{3}{8}{\tan }^2\theta }{2}\ge \sqrt{\frac{2}{3}{\cot }^2\theta \times \frac{3}{8}{\tan }^2\theta }$
$\frac{2}{3}{\cot }^2\theta +\frac{3}{8}{\tan }^2\theta \ge 1$
Therefore, $\frac{37}{24}+\frac{2}{3}{\cot }^2\theta +\frac{3}{8}{\tan }^2\theta \ge \frac{37}{24}+1=\frac{61}{24}$
Hence, option 'D' is correct.