By manipulating the left side of the equation, we will show that the expression \(1 + \tan^2 (t)\) is equivalent to \(\dfrac{1}{\cos^2 (t)}\text{.}\) First, we use the tangent identity to write the expression in terms of sines and cosines:
\begin{equation*}
1 + \tan^2 (t) = 1 +\left(\dfrac{\sin (t)}{\cos (t)}\right)^2 = 1 + \dfrac{\sin^2 (t)}{\cos^2 (t)}
\end{equation*}
Next, we notice that the right side of the proposed identity has only one term, so we combine the terms on the left side. So that the fractions have the same denominator, we write 1 as \(\dfrac{\cos^2 (t)}{\cos^2 (t)}\text{.}\)
\begin{equation*}
\alert{1} + \dfrac{\sin^2 (t)}{\cos^2 (t)} = \alert{\dfrac{\cos^2 (t)}{\cos^2 (t)}} + \dfrac{\sin^2 (t)}{\cos^2 (t)} = \dfrac{\cos^2 (t) + \sin^2 (t)}{\cos^2 (t)}
\end{equation*}
Finally, we apply the Pythagorean identity to the numerator.
\begin{equation*}
\dfrac{\cos^2 (t) + \sin^2 (t)}{\cos^2 (t)} = \dfrac{1}{\cos^2 (t)}
\end{equation*}
Thus, \(1 + \tan^2 (t) = \dfrac{1}{\cos^2 (t)}\text{,}\) and the identity is proved.