The expression (1 + x + ^2 x) (1 - x + ^2 x) has the positive val… - Vidyadip

MCQ

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 $
✓
For all $x \in R$
D
$x \ge 0$

✓

Answer

Correct option: C.

For all $x \in R$

c
(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 $x.$

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