Question:

<p><strong>Step 1:</strong> Apply sum-to-product formulas to the numerator and denominator.</p><p>For numerator: $\sin x + \sin 5x = 2\sin\left(\frac{x+5x}{2}\right)\cos\left(\frac{5x-x}{2}\right) = 2\sin(3x)\cos(2x)$</p><p>For denominator: $\cos x + \cos 5x = 2\cos\left(\frac{x+5x}{2}\right)\cos\left(\frac{5x-x}{2}\right) = 2\cos(3x)\cos(2x)$</p><p><strong>Step 2:</strong> Simplify the function by canceling common terms.</p><p>$$f(x) = \frac{2\sin(3x)\cos(2x)}{2\cos(3x)\cos(2x)} = \frac{\sin(3x)}{\cos(3x)} = \tan(3x)$$</p><p>(Note: This simplification is valid where $\cos(2x) \neq 0$)</p><p><strong>Step 3:</strong> Find the period of $\tan(3x)$.</p><p>The period of $\tan(kx)$ is $\frac{\pi}{k}$. Therefore, the period of $\tan(3x)$ is $\frac{\pi}{3}$.</p><p><strong>Step 4:</strong> Verify by checking periodicity.</p><p>$$f(x + \frac{\pi}{3}) = \tan\left(3\left(x + \frac{\pi}{3}\right)\right) = \tan(3x + \pi) = \tan(3x) = f(x)$$</p><p>This confirms the period is $\frac{\pi}{3}$. However, we must verify this works for the original function.</p><p><strong>Step 5:</strong> Check the original function's period more carefully.</p><p>Testing $f(x + \pi) = \frac{\sin(x+\pi) + \sin(5x+5\pi)}{\cos(x+\pi) + \cos(5x+5\pi)} = \frac{-\sin x - \sin 5x}{-\cos x - \cos 5x} = \frac{\sin x + \sin 5x}{\cos x + \cos 5x} = f(x)$</p><p>The fundamental period of the original function is $\pi$, as this is the smallest positive value where the simplified form equals the original.</p><p><strong>∴ Answer:</strong> C</p>