3. Given $\sin \alpha+\cos \alpha=\frac{7}{5}$, and $\tan \alpha>1$. Then $\boldsymbol{\operatorname { c o s }} \alpha=(\quad$. .
(A) $-\frac{3}{5}$
(B) $-\frac{4}{5}$
(C) $\frac{3}{5}$
(D) $\frac{4}{5}$
3.C.

Combining $\sin ^{2} \alpha+\cos ^{2} \alpha=1$, we easily get $\cos \alpha=\frac{3}{5}$.