False

Given, $a$ and $b$ are two distinct numbers such that $a b >0$.

Using, $AM > GM$

[since, $AM$ and $GM$ of two number $a$ and $b$ are $\frac{a+b}{2}$ and $\sqrt{a b}$, respectively]

$\Rightarrow \frac{a^{2}+b^{2}}{2}>\sqrt{a^{2} \cdot b^{2}}$

$\Rightarrow a^{2}+b^{2}>2 a b$

$\Rightarrow \frac{a^{2}+b^{2}}{2 a b}>1$ $\left[\because \cos \theta=\frac{a^{2}+b^{2}}{2 a b}\right]$

$\Rightarrow \cos \theta>1$ $[\because-1 \leq \cos \theta \leq 1]$

which is not possible.

Hence, $\cos \theta \neq \frac{a^{2}+b^{2}}{2 a b}$