Question

If $y=\tan 3x.\cot x$ then which of the following values is not possible for $y$?

• A. $-5$
• B. $-1$
• C. $1$
• D. $5$

Answer 1

Answer: (C)

$1$

$y=\tan 3x\cot x$

$⟹y=\frac{\tan 3x}{\tan x}$

$⟹\tan x\ne 0$

$⟹y=\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x}\times \frac{1}{\tan x}$ $(\tan 3\theta =\frac{3\tan x-{\tan }^{3}x}{1-3{\tan }^{2}x})$

$y=\frac{3-{\tan }^{2}x}{1-3{\tan }^{2}x}$

$y-3y{\tan }^{2}x=3-{\tan }^{2}x$

$(3y-1){\tan }^{2}x=y-3$

${\tan }^{2}x=\frac{y-3}{3y-1}>0$

$⟹\frac{y-3}{3y-1}>0$

$⟹(y-3)(y-\frac{1}{3})>0$

$y\in (-\infty ,\frac{1}{3})\cup (3,\infty )$