Question

The minimum value of $16\cot x+9\tan x$ in $(0,\frac{\pi }{2})$ is

• A. $12$
• B. $6$
• C. $24$
• D. $25$

Answer 1

Answer: (C)

$24$

In given interval, both $\cot x$ and $\tan x$ are positive.

Use $AM\ge GM$ for positive quantities.
$\frac{16\cot x+9\tan x}{2}\ge (16\cot x.9\tan x{)}^{\frac{1}{2}}$
$16\cot x+9\tan x\ge 24$