# How do you verify 1-cos(2x)*sec^2(x) = (tan(x))^2?

LHS=$1 - \cos \left(2 x\right) \cdot {\sec}^{2} \left(x\right)$
putting $\cos 2 x = 2 {\cos}^{2} x - 1$
$= 1 - \left(2 {\cos}^{2} x - 1\right) \cdot {\sec}^{2} \left(x\right)$
$= 1 - \left(2 {\cos}^{2} x \cdot {\sec}^{2} x - {\sec}^{2} \left(x\right)\right)$
$= 1 - 2 + {\sec}^{2} x = {\sec}^{2} x - 1 = {\tan}^{2} x = {\left(\tan x\right)}^{2} = R H S$