Question

If $A={\cos }^{2}x+\frac{1}{{\cos }^{2}x},B=\cos x-\frac{1}{\cos x}\forall x\ne (2n\pm 1)\frac{\pi }{2}$, then the minimum value of $\frac{A}{B}$ is

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $\frac{1}{\sqrt{2}}$
• D. None of these

Answer 1

Answer: (B)

$2\sqrt{2}$

$\frac{A}{B}=\frac{{\cos }^{2}x+\frac{1}{{\cos }^{2}x}}{\cos x-\frac{1}{\cos x}}=\frac{{(\cos x-\frac{1}{\cos x})}^2+2}{(\cos x-\frac{1}{\cos x})}$

$=(\cos x-\frac{1}{\cos x})+\frac{2}{(\cos x-\frac{1}{\cos x})}$
$\frac{A}{B}=B+\frac{2}{B}\ge 2\sqrt{2}$