Trigonometric Equations Question 56
Question: The expression $ (1+\tan x+{{\tan }^{2}}x) $ $ (1-\cot x+{{\cot }^{2}}x) $ has the positive values for x, given by
Options:
A) $ 0\le x\le \frac{\pi }{2} $
B) $ 0\le x\le \pi $
C) For all $ x\in R $
D) $ x\ge 0 $
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Answer:
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 x $ . Hence it is positive for all value of x.
BETA
