9. If $\sin \theta+\cos \theta=\frac{7}{5}$, and $\tan \theta<1$, then $\sin \theta=$ $\qquad$ .
9. $\frac{3}{5}$.

Squaring both sides of $\sin \theta+\cos \theta=\frac{7}{5}$, we get
$$
\begin{array}{l}
2 \sin \theta \cdot \cos \theta=\frac{24}{25} \\
\Rightarrow(\sin \theta-\cos \theta)^{2}=1-2 \sin \theta \cdot \cos \theta=\frac{1}{25} \\
\Rightarrow \sin \theta-\cos \theta= \pm \frac{1}{5} \\
\Rightarrow\left\{\begin{array} { l } 
{ \sin \theta + \cos \theta = \frac { 7 } { 5 } , } \\
{ \sin \theta - \cos \theta = \frac { 1 } { 5 } }
\end{array} \text { or } \left\{\begin{array}{l}
\sin \theta+\cos \theta=\frac{7}{5}, \\
\sin \theta-\cos \theta=-\frac{1}{5}
\end{array}\right.\right. \\
\Rightarrow\left\{\begin{array} { l } 
{ \sin \theta = \frac { 4 } { 5 } , } \\
{ \cos \theta = \frac { 3 } { 5 } }
\end{array} \text { or } \left\{\begin{array}{l}
\sin \theta=\frac{3}{5}, \\
\cos \theta=\frac{4}{5} .
\end{array}\right.\right.
\end{array}
$$

Since $\tan \theta<1$, we have $\sin \theta=\frac{3}{5}$.