Given \( \sec x = \frac{5}{4} \), then \( \cos x = \frac{4}{5} \). Using:
\[\n\tan^2 x = \sec^2 x - 1\n\]
Substitute \( \sec x \):
\[\n\tan^2 x = \left(\frac{5}{4}\right)^2 - 1 = \frac{25}{16} - 1 = \frac{9}{16}.\n\]
Thus, \( \tan x = \frac{3}{4} \). Using:
\[\n\frac{\tan x}{1 + \tan^2 x} = \frac{\frac{3}{4}}{1 + \frac{9}{16}} = \frac{\frac{3}{4}}{\frac{25}{16}} = \frac{3}{4} \times \frac{16}{25} = \frac{9}{25}.\n\]
The answer is \( \frac{9}{25} \).