Answer
$y' = 1 +tan^{2}x=sec^2{x}$
Work Step by Step
$y = (sinx +cosx)secx$
$y' = (cosx-sinx)secx + secx.tanx(sinx+cosx)$
$y' = cosx\ secx - sinx\ secx + secx\ tanx\ sinx + secx\ tanx\ cosx$
$y' = 1 - tanx +tan^{2}x + tanx$
$y' = 1 +tan^{2}x$