\(\cos\theta=\dfrac{a^2+b^2}{2ab}\), where \(a,b\) are distinct and \(ab\gt0\).
False.
Step 1: Recall that for any angle \(\theta\), the value of cosine is always between -1 and 1.
That is: \(-1 \leq \cos\theta \leq 1\).
Step 2: Now look at the right-hand side: \(\dfrac{a^2+b^2}{2ab}\).
Step 3: Use the AM–GM inequality:
For any positive numbers \(a\) and \(b\),
\(\dfrac{a^2+b^2}{2} \geq ab\),
and equality holds only if \(a=b\).
Step 4: Divide both sides by \(ab\gt0\):
\(\dfrac{a^2+b^2}{2ab} \geq 1\),
with equality only if \(a=b\).
Step 5: But in our question, \(a\) and \(b\) are distinct (not equal).
So, \(\dfrac{a^2+b^2}{2ab} \gt 1\).
Step 6: This means the RHS is always greater than 1.
But cosine values cannot be greater than 1.
Therefore, the given relation cannot be true.
Final Conclusion: The statement is False.