Answer
$\frac{2}{3}.$
Work Step by Step
$$\lim_{x\to \frac{\pi}{4}}\frac{\tan{x}}{1+\cos^2{x}}\\=\frac{\tan{\frac{\pi}{4}}}{1+\cos^2{\frac{\pi}{4}}}\\=\frac{1}{1+(\frac{\sqrt2}{2})^2}\\=\frac{1}{1+\frac{1}{2}}\\=\frac{1}{\frac{3}{2}}=\frac{2}{3}.$$
