Ta có:$\star$ $\tan^2 \alpha +1=\frac{1}{\cos^{2} \alpha}\Leftrightarrow \sqrt{\tan^2 \alpha +1}=\frac{1}{\cos \alpha}$
$\star$ $\sin \alpha + \cos \alpha = \frac{7}{5}$
$\Leftrightarrow \tan \alpha +1 = \frac{7}{5\cos \alpha}$ (chia 2 vế cho $\cos \alpha$)
$\Leftrightarrow \tan \alpha +1 = \frac{7}{5}\sqrt{\tan^2 \alpha +1}$
$\Leftrightarrow 25(\tan \alpha +1)^2=49(\tan^2 \alpha +1)$
$\Leftrightarrow 24 \tan^{2} \alpha -50 \tan \alpha +24=0$
$\Leftrightarrow \left[ \begin{array}{l} \tan \alpha = \frac{3}{4}\\ \tan \alpha = \frac{4}{3} \end{array} \right.$