Question

If $x\in (0,\frac{\pi }{2})$ the minimum value of the expression $(1+\tan x+{\tan }^{2}x)(1-\cot x+{\cot }^{2}x)$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Given $y=(1+tanx+{tan}^{2}x)(1-cotx+{cot}^{2}x)$

$y=(1+tanx+{tan}^{2}x)(1-\frac{1}{tanx}+\frac{1}{{tan}^{2}x})$

Put $tanx=u$

$\therefore x={tan}^{-1}u$

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

$\therefore y=1-\frac{1}{u}+\frac{1}{u^2}+u-1+\frac{1}{u}+u^2-u+1$

$\therefore y=\frac{1}{u^2}+u^2+1$

For y to be maximum or minimum, differentiate y w.r.t. x and equate to zero.

$\therefore \frac{dy}{du}=\frac{d}{du}[\frac{1}{u^2}+u^2+1]=0$

$\therefore \frac{-2}{u^3}+2u=0$

$\therefore 2u=\frac{2}{u^3}$

$\therefore u=\frac{1}{u^3}$

$\therefore u^4=1$

$\therefore u=1$

Thus, minimum value of expression is 1.