A) \[0\le x\le \frac{\pi }{2}\]
B) \[0\le x\le \pi \]
C) For all \[x\in R\]
D) \[x\ge 0\]
Correct Answer: C
Solution :
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, \[1+{{\tan }^{2}}x\ge {{\tan }^{2}}x,\text{ }\forall \text{ }x\]. Hence it is positive for all value of x.
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