Question

The expression $\frac{\tan (x+\alpha )}{\tan (x-\alpha )}$ cannot lie between

• A. ${\tan }^2\left(\frac{\pi }{4}-\alpha \right)$ and ${\tan }^2\left(\frac{\pi }{4}+a\right)$
• B. ${\sin }^2\left(\frac{\pi }{4}-\alpha \right)$ and ${\sin }^2\left(\frac{\pi }{4}+\alpha \right)$
• C. ${\cos }^2\left(\frac{\pi }{4}-\alpha \right)$ and ${\cos }^2\left(\frac{\pi }{4}+\alpha \right)$
• D. ${\tan }^2\alpha$ and ${\cos }^2\left(\frac{\pi }{4}+\alpha \right)$

Answer 1

The correct option is A ${\tan }^2\left(\frac{\pi }{4}-\alpha \right)$ and ${\tan }^2\left(\frac{\pi }{4}+a\right)$

Let $y=\frac{\tan (x+\alpha )}{\tan (x-\alpha )}$

$y=\frac{(\tan x+\tan \alpha \left)\right(1+\tan x\tan \alpha )}{(\tan x-\tan \alpha )(1-\tan x\tan \alpha )}$

Put $\tan x=u$ and $\tan \alpha =c$

$y=\left(\frac{u+c}{1-cu}\right)\left(\frac{1+cu}{u-c}\right)$

$y=\frac{c{u}^2+(1+c^2)u+c}{-c{u}^2+(1+c^2)u-c}$

$\Rightarrow c(y+1)u^2+(1+c^2)(1-y)u+c(1+y)=0$ .....(1)

Since, $u$ is real $D\ge 0$

$(1+c^2{)}^2(1-y{)}^2-4c^2(1+y{)}^2\ge 0$

$\Rightarrow (1-c^2{)}^2{y}^2-2[(1+c^2{)}^2+4c^2]y+(1-c^2{)}^2\ge 0$ ....(2)

Discriminant of equation (2)

$D=4\left[\right(1+c^2{)}^2+4c^2{]}^2-4\left[\right(1-c^2{)}^2{]}^2$

$D=64(1+c^2{)}^2{c}^2$.

Roots of equation (2)
$=\frac{2\left[\right(1+c^2{)}^2+4c^2]\pm 8(1+c^2)c}{2(1-c^2{)}^2}$

$=\frac{(1+c^2{)}^2+4c^2\pm 4c(1+c^2)}{(1-c^2{)}^2}$

$=\frac{(1+c^2\pm 2c{)}^2}{(1-c^2{)}^2}=\frac{(1\pm c{)}^4}{(1-c^2{)}^2}=(\frac{1-c}{1+c}{)}^2,(\frac{1+c}{1-c}{)}^2$

$={\left(\frac{1-\tan \alpha }{1+\tan \alpha }\right)}^2,{\left(\frac{1+\tan \alpha }{1-\tan \alpha }\right)}^2$

$\therefore {\tan }^2(\frac{\pi }{4}-\alpha ),{\tan }^2(\frac{\pi }{4}+\alpha )$

Since roots of equation (2) are real and unequal and coefficient of $y^2=(1-c^2{)}^2>0$

$\therefore (1-c^2{)}^2.y^2-2[(1+c^2{)}^2+4c^2]y+(1-c^2{)}^2\ge 0$

$\Rightarrow y$ does not lie between

${\tan }^2(\frac{\pi }{4}-\alpha )$ and ${\tan }^2(\frac{\pi }{4}+\alpha )$

Hence, option A.