Question

The value of tan $3\alpha cot\alpha$ cannot lie in

• A. $[0,2/3]$
• B. $(1/3,3)$
• C. $[4/3,4]$
• D. $[2,10/3]$

Answer 1

The correct option is B $(1/3,3)$

Consider $\tan 3\alpha \cot \alpha$
$=\frac{3\tan \alpha -{\tan }^3\alpha }{\tan \alpha (1-3{\tan }^2\alpha )}$
$=\frac{3-{\tan }^2\alpha }{1-3{\tan }^2\alpha }=x\left(say\right)$
$\Rightarrow {\tan }^2\alpha =\frac{x-3}{3x-1}=\frac{(3x-1)(x-3)}{(3x-1{)}^2}$
Since $ta{n}^2\alpha \ge 0$

$\frac{(3x-1)(x-3)}{(3x-1{)}^2}\ge 0$

$\Rightarrow (3x-1)(x-3)\ge 0$

$\Rightarrow x\le \frac{1}{3}$ or $x\ge 3$
So $x$ cannot lie in $(\frac{1}{3},3)$