$\underset{y→0}{\lim}\frac{(x+y)\sec(x+y)-x\sec x}{y}$ is equal to

$\sec x(x\tan x+1)$

$\underset{y→0}{\lim}\begin{Bmatrix}\frac{x\{sec(x+y)-\sec x\}}{y}+\sec(x+y)\end{Bmatrix}$

$\underset{y→0}{\lim}\begin{bmatrix}\frac{x}{y}\begin{Bmatrix}\frac{\cos x-\cos(x+y)}{\cos(x+y)\cos x}\end{Bmatrix}\end{bmatrix}+\underset{y→0}{\lim}\sec(x+y)$

$=\underset{y→0}{\lim}\begin{bmatrix}\frac{x\sin(x+\frac{y}{2})}{\cos(x+y).\cos x}.\frac{\sin(\frac{y}{2})}{\frac{y}{2}}\end{bmatrix}+\sec x=x\tan x\sec x+\sec x=\sec x(x\tan x+1)$