We are given that \(\cos(\theta) = \frac{a}{b}\). To find \(1 + \tan^2(\theta)\), we can use the trigonometric identity:

\(\tan^2(\theta) = \sec^2(\theta) - 1\)

And since \(\sec(\theta) = \frac{1}{\cos(\theta)}\), we can substitute the value of \(\cos(\theta)\) into the identity:

\(\tan^2(\theta) = \left(\frac{1}{\cos(\theta)}\right)^2 - 1 = \frac{1}{\cos^2(\theta)} - 1\)

Now, we know that \(\cos(\theta) = \frac{a}{b}\), so we can substitute it into the equation:

\(\tan^2(\theta) = \frac{1}{\left(\frac{a}{b}\right)^2} - 1 = \frac{1}{\frac{a^2}{b^2}} - 1\)

Now, to simplify further, we can divide 1 by a fraction by multiplying it by the reciprocal of the fraction:

\(\tan^2(\theta) = 1 \cdot \frac{b^2}{a^2} - 1 = \frac{b^2}{a^2} - 1\)

Finally, add 1 to this expression:

\(1 + \tan^2(\theta) = 1 + \left(\frac{b^2}{a^2} - 1\right)\)

Now, combine like terms:

\(1 + \tan^2(\theta) = \frac{b^2}{a^2} - 1 + 1\)

The -1 and +1 cancel out, leaving us with:

\(1 + \tan^2(\theta) = \frac{b^2}{a^2}\)

So, \(1 + \tan^2(\theta) = \frac{b^2}{a^2}\)