Write 'True' or 'False' and justify your answer.
$\cos \theta = \frac{a^{2} + b^{2}}{2ab}$,where $a$ and $b$ are two distinct numbers such that $ab > 0$.

(B) False.
Given that $a$ and $b$ are two distinct numbers such that $ab > 0$.
We know that for any two distinct positive numbers,the Arithmetic Mean $(AM)$ is strictly greater than the Geometric Mean $(GM)$.
$AM > GM$
$\frac{a^2 + b^2}{2} > \sqrt{a^2 b^2}$
$\frac{a^2 + b^2}{2} > ab$
Dividing both sides by $ab$ (since $ab > 0$):
$\frac{a^2 + b^2}{2ab} > 1$
Since $\cos \theta = \frac{a^2 + b^2}{2ab}$,this implies $\cos \theta > 1$.
However,the range of $\cos \theta$ is $[-1, 1]$,meaning $\cos \theta$ cannot be greater than $1$.
Therefore,the given statement is False.