If \(\cos A=\frac{12}{13}\)
Then \(\cos ^{2} A=\left(\frac{12}{13}\right)^{2}=\frac{144}{169}\)
from \(\sin ^{2} A+\cos ^{2} A=1\)
\begin{array}{l}
\sin ^{2} A=1-\cos ^{2} A=1-\frac{144}{169} \\
\sin ^{2} A=\frac{25}{169} \\
\tan ^{2} A=\frac{\sin ^{2} A}{\cos ^{2} A}=\frac{25}{144} 169
\end{array}
\(\tan ^{2} A=\frac{25}{169} \times \frac{169}{144}=\frac{25}{144}\)
\begin{array}{l}
1+\tan ^{2} A=1+\frac{25}{144}=\frac{144+25}{144} \\
1+\tan ^{2} A=\frac{169}{144} .
\end{array}