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

Given:

\( \cos \theta=\frac{a^{2}+b^{2}}{2 a b} \), where \( a \) and \( b \) are two distinct numbers such that \( a b>0 \).

To do:

We have to find whether the given statement is true or false.

Solution:

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

This implies,

$AM>GM$

AM and GM of two number $a$ and $b$ are $\frac{a+b}{2}$ and $\sqrt{a b}$

Therefore,

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

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

$\frac{a^{2}+b^{2}}{2 a b}>1$

$\cos \theta=\frac{a^{2}+b^{2}}{2 a b}$

$\cos \theta>1$ which is not possible.        [Since $-1 \leq \cos \theta \leq 1$]

Hence,

$\cos \theta≠\frac{a^{2}+b^{2}}{2 a b}$.