G9.1 If $\sin \theta=\frac{3}{5}, a=\sqrt{\tan ^{2} \theta+1}$, find $a$.
$\begin{array}{l}\cos ^{2} \theta=1-\sin ^{2} \theta=1-\frac{9}{25}=\frac{16}{25} \\ \tan ^{2} \theta=\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{\frac{9}{25}}{\frac{16}{25}}=\frac{9}{16} \\ a=\sqrt{\tan ^{2} \theta+1}=\sqrt{\frac{9}{16}+1}=\frac{5}{4}\end{array}$