Answer
$$\ln (2)$$
Work Step by Step
$$Area=\int_{0}^{\pi/2} (\sec x- \tan x) dx \\ = \lim\limits_{u \to \dfrac{\pi}{2}^{-}} \int_0^u (\sec x- \tan x) dx \\=\lim\limits_{u \to \dfrac{\pi}{2}^{-}}[\ln |\dfrac{(\sec x+ \tan x)}{\sec x}|]_0^u \\=\lim\limits_{u \to \dfrac{\pi}{2}^{-}}[\ln |1+\sin x|]_0^u \\=[\ln |1+\sin x|]_0^{\pi/2} \\=\ln |1+\sin ( \dfrac{\pi}{2})|-\ln |1+\sin (0)| \\=\ln 2$$
