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