A) \[x\in (0,\pi /2)\]
B) \[x\in (-\pi /2,0)\]
C) \[x\in R\]
D) None of these
Correct Answer: D
Solution :
[d] \[1+{{\tan }^{2}}x-\sec x\tan x<0\] \[\Rightarrow \,\,\,\,{{\sec }^{2}}x-\sec x\tan x<0\] \[\Rightarrow \,\,\,\,\,\sec x(\sec x-\tan x)<0\] \[\Rightarrow \,\,\,\frac{\sec x}{\sec x+\tan x}<0\] \[\Rightarrow \,\,\,\frac{1}{1+\sin x}<0;\]This is not possible
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