Question

Let $\frac{\sin x}{a}=\frac{\cos x}{b}=\frac{\tan x}{c}=2,$ where $0<x<\frac{\pi }{2}$ and $R=bc+\frac{1}{2c}+\frac{2a}{1+2b}.$ Then the minimum value of $R$ is

• A. $1$
• B. $2$
• C. $-1$
• D. $4$

Answer 1

The correct option is A $1$

$\frac{\sin x}{a}=\frac{\cos x}{b}=\frac{\tan x}{c}=2$
$\therefore a=\frac{\sin x}{2},b=\frac{\cos x}{2},c=\frac{\tan x}{2}$
We have $R=bc+\frac{1}{2c}+\frac{2a}{1+2b}$
$\Rightarrow R=\frac{\cos x\cdot \tan x}{4}+\frac{1}{\tan x}+\frac{\sin x}{1+\cos x}$
$=\frac{\sin x}{4}+\frac{\cos x}{\sin x}+\frac{\sin x}{1+\cos x}$
$=\frac{\sin x}{4}+\frac{\cos x(1+\cos x)+{\sin }^{2}x}{\sin x(1+\cos x)}$
$=\frac{\sin x}{4}+\frac{1}{\sin x}$

Now, by A.M. $\ge$ G.M.
$\frac{1}{2}(\frac{\sin x}{4}+\frac{1}{\sin x})\ge \sqrt{\frac{1}{4}\sin x\cdot \frac{1}{\sin x}}$
$\therefore \frac{\sin x}{4}+\frac{1}{\sin x}\ge 1$
So, the minimum value of $R$ is $1.$