The expression (1 + tan x + tan^2 x)(1 - cot | vedclass

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(C) Let the expression be $E = (1 + 	an x + 	an^2 x)(1 - \cot x + \cot^2 x)$.Substituting $\cot x = \frac{1}{	an x}$,we get $E = (1 + 	an x + 	an

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For all $x \in \mathbb{R}$

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(C) Let the expression be $E = (1 + 	an x + 	an^2 x)(1 - \cot x + \cot^2 x)$.Substituting $\cot x = \frac{1}{	an x}$,we get $E = (1 + 	an x + 	an^2 x)(1 - \frac{1}{	an x} + \frac{1}{	an^2 x})$.$E = (1 + 	an x + 	an^2 x) \left( \frac{	an^2 x - 	an x + 1}{	an^2 x} ight)$.$E = \frac{(1 + 	an^2 x + 	an x)(1 + 	an^2 x - 	an x)}{	an^2 x}$.Using the identity $(a+b)(a-b) = a^2 - b^2$,where $a = 1 + 	an^2 x$ and $b = 	an x$:$E = \frac{(1 + 	an^2 x)^2 - 	an^2 x}{	an^2 x}$.Since $1 + 	an^2 x \ge 2|	an x|$ by $AM$-$GM$ inequality,$(1 + 	an^2 x)^2 > 	an^2 x$ for all $x$ where $	an x$ is defined.Thus,the expression is positive for all $x \in \mathbb{R}$ except where $	an x$ is undefined or zero.

The expression $(1 + \tan x + \tan^2 x)(1 - \cot x + \cot^2 x)$ has positive values for $x$,given by

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