Question

The maximum value of $\frac{4}{{\sec }^{2}x+{\text{cosec}}^{2}x}$ is

Answer 1

$\frac{4}{{\sec }^{2}x+{\text{cosec}}^{2}x}=\frac{4}{\frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}}=\frac{4{\cos }^{2}x{\sin }^{2}x}{{\cos }^{2}x+{\sin }^{2}x}={\sin }^{2}2x$
So, maximum value $=1$