If sec x=t

Given sec $x=t+\frac{1}{4 t}$

Since $\tan ^{2} x=\sec ^{2} x-1$

$=\left(t+\frac{1}{4 t}\right)^{2}-1$

$=t^{2}+\frac{1}{16 t^{2}}+\frac{1}{2}-1$

$=t^{2}+\frac{1}{16 t^{2}}-\frac{1}{2}$

$\tan ^{2} x=\left(t-\frac{1}{4 t}\right)^{2}$

i. e $\tan x=\pm\left(t-\frac{1}{4 t}\right)$

$\therefore \sec x+\tan x=\left(t+\frac{1}{4 t}\right) \pm\left(t-\frac{1}{4 t}\right)$

Hence, sec $x+\tan x=2 t$ or $\frac{1}{2 t}$