7.4 Given $\sin \alpha-\cos \alpha=\frac{1}{5}, 0<\alpha<180^{\circ}$. Then the value of $\operatorname{tg} \alpha$ is
(A) $\frac{3}{4}$.
(B) $-\frac{3}{4}$.
(C) $\frac{4}{3}$.
(D) $-\frac{4}{3}$.
(Taiyuan, Shanxi Province, China Junior High School Mathematics Competition, 1996)
[Solution] $\because \sin \alpha-\cos \alpha=\frac{1}{5}, \therefore \quad(\sin \alpha-\cos \alpha)^{2}=\frac{1}{25}$,
that is
$$
\begin{array}{l}
1-2 \sin \alpha \cos \alpha=\frac{1}{25}, 2 \sin \alpha \cos \alpha=\frac{24}{25}, 1+2 \sin \alpha \cos \alpha=\frac{49}{25}, \\
\therefore \quad(\sin \alpha+\cos \alpha)^{2}=\frac{49}{25}, \sin \alpha+\cos \alpha= \pm \frac{7}{5} . \\
\text { From }\left\{\begin{array} { l } 
{ \sin \alpha - \cos \alpha = \frac { 1 } { 5 } , } \\
{ \sin \alpha + \cos \alpha = \frac { 7 } { 5 } ; }
\end{array} \text { we get } \left\{\begin{array}{l}
\sin \alpha=\frac{4}{5}, \\
\cos \alpha=\frac{3}{5} .
\end{array}\right.\right. \\
\therefore \quad \tan \alpha=\frac{4}{3} . \\
\text { From }\left\{\begin{array} { l } 
{ \sin \alpha - \cos \alpha = \frac { 1 } { 5 } , } \\
{ \sin \alpha + \cos \alpha = - \frac { 7 } { 5 } ; }
\end{array} \text { we get } \left\{\begin{array}{l}
\sin \alpha=-\frac{3}{5}, \\
\cos \alpha=-\frac{4}{5} .
\end{array}\right.\right.
\end{array}
$$
$$
\because 0<\alpha<180^{\circ} \text {, this solution is discarded. }
$$

Therefore, the answer is $(C)$.