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 :
\[(1+\tan x+{{\tan }^{2}}x)(1-\cot \,x+{{\cot }^{2}}x)\] \[=\frac{(1+\tan x+{{\tan }^{2}}x)(1+{{\tan }^{2}}x-\tan x)}{{{\tan }^{2}}x}\] \[=\frac{{{(1+{{\tan }^{2}}x)}^{2}}-{{\tan }^{2}}x}{{{\tan }^{2}}x}\] Obviously, \[1+{{\tan }^{2}}x\,\,\,\ge \,\,{{\tan }^{2}}x,\,\forall x\in R\]
You need to login to perform this action.
You will be redirected in
3 sec
