Question

The minimum value of $4{\text{cosec}}^{2}x+3{\sec }^{2}x+2$ is

• A. $9+2\sqrt{3}$
• B. $9-2\sqrt{3}$
• C. $9+4\sqrt{3}$
• D. $9-4\sqrt{3}$

Answer 1

The correct option is C $9+4\sqrt{3}$

$4{\text{cosec}}^{2}x+3{\sec }^{2}x+2=4(1+{\cot }^{2}x)+3(1+{\tan }^{2}x)+2=4+4{\cot }^{2}x+3+3{\tan }^{2}x+2=9+4{\cot }^{2}x+3{\tan }^{2}x$

We know that, $a+b\ge 2\sqrt{ab}$ if $a,b>0$

Let $a=4{\cot }^{2}x,b=3{\tan }^{2}x$
$4{\cot }^{2}x+3{\tan }^{2}x\ge 4\sqrt{3}\cdot \sqrt{{\cot }^{2}x\cdot {\tan }^{2}x}\Rightarrow 9+4{\cot }^{2}x+3{\tan }^{2}x\ge 9+4\sqrt{3}$