Answer
This set $is$ linearly dependent in $\mathbb{P}_3$.
Work Step by Step
Two vectors are linearly dependent if and only if one vector is a scalar multiple of the other. However, it is clear that neither of $1+t^2$ and $1-t^2$ is a scalar multiple of the other, because their constant terms are equal, so attempting to scale one so that they have equal $t^2$ terms will disrupt the equality of the constant terms.
Alternatively, suppose $$c_1(1+t^2)+c_2(1-t^2)=(c_1+c_2)+(c_1-c_2)t^2=0$$ for some scalars $c_1$ and $c_2$ and for all real values of $t$. Letting $t=0$, we see that $c_ + c_2=0$. Now we can differentiate both sides of the equation $(c_1+c_2)+(c_1-c_2)t^2=0$ twice, yielding $2(c_1-c_2)=0$, which implies $c_1-c_2=0$. But the system of equations
\begin{cases}
c_1 + c_2 = 0 \\
c_1 - c_2 = 0
\end{cases} has only one solution, namely, $c_1 = c_2 = 0$. Therefore, $1+t^2$ and $1-t^2$ are linearly independent.
