If set A = \ x: tan x = sec x, x ∈ [0,4π]\ and set B = \ x: ^ 2 x =...

For $A, \dfrac{sin x}{cos x} = \dfrac{1}{cos x} ⇒ sin x = 1$
$∴ x = \dfrac{\pi}{2}, \dfrac{5\pi}{2}$
but at $x = \dfrac{\pi}{2}, \dfrac{5\pi}{2}$

$tan x, sec \theta$ is not defined
$∴ A = ϕ$
for $B, sin x = 1, sin x = - 1$
$x = \dfrac{\pi}{2}, \dfrac{5\pi}{2}orx = \dfrac{3\pi}{2}, \dfrac{7\pi}{2}$
$∴ B = \left\{\dfrac{\pi}{2}, \dfrac{3\pi}{2}, \dfrac{5\pi}{2}, \dfrac{7\pi}{2}\right\}$
$∴ n\left(A \times B\right) = 0$