If sec x

Given $\sec x-\tan x=\frac{2}{3}$ .....(1)

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

$\Rightarrow(\sec x-\tan x)(\sec x+\tan x)=1$

$\Rightarrow \frac{2}{3}(\sec x+\tan x)=1$

$\Rightarrow \sec x+\tan x=\frac{3}{2}$   ....(2)

Subtracting (1) from (2)

$\sec x+\tan x-\sec x+\tan x=\frac{3}{2}-\frac{2}{3}$

$\Rightarrow 2 \tan x=\frac{9-4}{6}$

$\Rightarrow 2 \tan x=\frac{5}{6}$

$\Rightarrow \tan x=\frac{5}{12}$