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We are given the function \(f(x) = a(\tan x - \cot x)\), where \(a > 0\). We can rewrite \(\cot x\) as \(\frac{1}{\tan x}\), so the function becomes: $$f(x) = a\left(\tan x - \frac{1}{\tan x}\right)$$
To determine whether the function is increasing or decreasing, we need to find its derivative, \(f'(x)\). $$f'(x) = a\left(\frac{d}{dx}(\tan x) - \frac{d}{dx}(\cot x)\right)$$ We know that \(\frac{d}{dx}(\tan x) = \sec^2 x\) and \(\frac{d}{dx}(\cot x) = -\csc^2 x\). Therefore, $$f'(x) = a(\sec^2 x - (-\csc^2 x)) = a(\sec^2 x + \csc^2 x)$$
Since \(a > 0\), we need to determine the sign of \(\sec^2 x + \csc^2 x\). We know that \(\sec^2 x\) and \(\csc^2 x\) are always non-negative (since they are squares). Also, they cannot both be zero at the same time. Therefore, \(\sec^2 x + \csc^2 x > 0\) for all \(x\) in the domain of \(f(x)\). Since \(a > 0\) and \(\sec^2 x + \csc^2 x > 0\), we have \(f'(x) = a(\sec^2 x + \csc^2 x) > 0\).
Since \(f'(x) > 0\) for all \(x\) in the domain of \(f(x)\), the function \(f(x)\) is an increasing function in its domain.
Final Answer: Increasing function