A) \[p\le 2\]
B) \[p\ge 2\]
C) \[p<2\]
D) \[p>2\]
Correct Answer: B
Solution :
Given, \[p={{\tan }^{2}}x+{{\cot }^{2}}x={{(\tan x+\cot x)}^{2}}-2\] \[={{\left( \frac{{{\sin }^{2}}x+{{\cos }^{2}}x}{\sin x\cos x} \right)}^{2}}-2={{\left( \frac{2}{\sin 2x} \right)}^{2}}-2=\frac{4}{{{\sin }^{2}}2x}-2\]The maximum value of sin \[2x\] is 1. \[\therefore \] \[{{p}_{\min }}=\frac{4}{1}-2=2\therefore p\ge 2\] Hence, \[p\ge 2.\]
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