Answer
Both functions have the same derivative, $2sec^2(x)tan(x)$.
This leads to the conclusion that $f(x)$ and $g(x)$ are the same function, but with different vertical shifts. That is:
$sec^2(x) = tan^2(x) + C$, where C is a constant.
Work Step by Step
The derivatives:
$f(x) = tan^2(x)$
$f'(x) = 2tan(x)*sec^2(x)$
$f'(x) = 2sec^2(x)tan(x)$
$g(x) = sec^2(x)$
$g'(x) = 2sec(x)*sec(x)tan(x)$
$g'(x) = 2sec^2(x)tan(x)$
Conclusion:
Since $2sec^2(x)tan(x)$ is the derivative of both $f(x)$ and $g(x)$, this must mean that the antiderivative of $2sec^2(x)tan(x)$ is both $f(x) + C$ and $g(x) + C$ by the First Fundamental Theorem of Calculus.
Therefore, since $\int2sec^2(x)tan(x)dx = f(x) + C_1 = g(x) + C_2$, $f(x)$ and $g(x)$ must be vertical shifts of one another.
