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  • cos theta = a power 2 + b power 2 by 2ab where a and b are two distinct numbers such that ab> 0.

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

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best_answer

We know that

-1\leq \cos \theta \leq 1
We also know that

\left ( a-b \right )^{2}= a^{2}+b^{2}-2ab
\text{Since }  \left ( a-b \right )^{2} \text{is a square term hence it is always positive }
\left ( a-b \right )^{2}> 0
a2 + b2 – 2ab > 0
a^{2}+b^{2}> 2ab
We observe that a^{2}+b^{2} is always greater than 2ab.
Hence,         

  \frac{a^{2}+b^{2}}{2ab}> 1
Because if we divide a big term by small then the result is always greater than 1.
cos\theta is always less than or equal to 1
Hence the given statement is false.

Posted by

infoexpert27

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