Expression (1 + tan x + {tan ^2}x) (1 - cot x + {cot ^2}x) has positive values for x , given by

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Trigonometrical Equations

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

A

$0 \le x \le \frac{\pi }{2}$

B

$0 \le x \le \pi $

C

For all $x \in R$

D

$x \ge 0$

Solution

(c) The expression is $\frac{{(1 + \tan x + {{\tan }^2}x)(1 + {{\tan }^2}x – \tan x)}}{{{{\tan }^2}x}}$

$= \frac{{{{(1 + {{\tan }^2}x)}^2} – {{\tan }^2}x}}{{{{\tan }^2}x}}$

Obviously, $1 + {\tan ^2}x \ge {\tan ^2}x,{\rm{ }}\forall {\rm{ }}x$.

Hence it is positive for all value of $x.$

Standard 11

Mathematics

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