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

Answer

False.Consider two numbers $a^2$ and $b^2$ then
Arithmetic mean (AM) of $a^2$ and $b^2=\frac{a^2+b^2}{2}$
Geometric mean (GM) of $a^2$ and $b^2=\sqrt{a^2 \times b^2}$
$\Rightarrow\ \text{GM}=\text{ab}$
$\because\ \text{AM}>\text{GM}$
$\therefore\ \frac{\text{a}^2+\text{b}^2}{2}>\text{ab}$
$\Rightarrow\ \frac{\text{a}^2+\text{b}^2}{2\text{ab}}>1$
$\Rightarrow\ \cos\theta>1\ \Big(\text{Given}\cos\theta=\frac{\text{a}^2+\text{b}^2}{2\text{ab}}\Big)$
But, the value of $\cos\theta$ can never be greater than 1.
So,the given expression is false.

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