If α and β are 2 roots of $a \tan θ+b \sec θ=c$ then how the roots of $(a^2-b^2)\tan^2θ-2ac \tanθ+(c^2-b^2)=0$ are $\tan α$ and $\tan β$?

Question

given that $a \tanθ+b \secθ=c$ I am told to prove that $\tan(α+β)=\frac{2ac}{a^2-c^2}$.

Now, $$a \tanθ+b \secθ=c...(1)$$$$(b \sec θ)^2=(c-a \tan θ)^2$$$$(a^2-b^2)\tan^2θ-2ac \tan θ+(c^2-b^2)=0...(2)$$ Now solution book says that if the initial equation(1) has the roots α and β then the final equation (2) will have the roots $\tan α$ and $\tan β$. I don't get this assumption. Just because we have squared the expression, why will the roots turn into tangent of the previous ones and not anything else?