Question

Let $y={(\sin x+\csc x)}^2+{(\cos x+\sec x)}^2$,then minimum value of y is-

• A. 7
• B. 3
• C. 9
• D. 0

Answer 1

The correct option is C 9

$y={(\sin x+\csc x)}^2+{(\cos x+\sec x)}^2={\sin }^{2}x+{\csc }^{2}x+2\sin x\csc x+{\cos }^{2}x+{\sec }^{2}x+2\cos x\sec x$
$=5+\frac{1}{{\sin }^{2}x{\cos }^{2}x}=5+\frac{4}{{\sin }^{2}2x}$
AS $0\le {\sin }^{2}2x\le 1$
Hence $min(5+\frac{4}{{\sin }^{2}2x})=9$