Prove the following


$\cos \theta=\frac{a^{2}+b^{2}}{2 a b}$ where $\mathrm{a}$ and $\mathrm{b}$ are two distinct numbers such that $\mathrm{ab}>0$.



Given, a and b are two distinct numbers such that ab > 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, $\cos \theta \neq \frac{a^{2}+b^{2}}{2 a b}$

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