• question_answer If $f(x)={{\cos }^{2}}x+{{\sec }^{2}}x,$ then  [MNR 1986] A) $f(x)<1$ B) $f(x)=1$ C) $1<f(x)<2$ D) $f(x)\ge 2$

Since ${{\left( x-\frac{1}{x} \right)}^{2}}\ge 0,\,\,\text{}\,\,x\in R,$ we have ${{x}^{2}}+\frac{1}{{{x}^{2}}}\ge 2$ and Hence, $f(x)={{\cos }^{2}}x+\frac{1}{{{\cos }^{2}}x}\ge 2$.