91Ó°ÊÓ

For \(x>0\), we have \(|x| = x\) and \(|\tan x| = \tan x\). It is a well-known fact that \(x \leq \tan x\) for \(0 \leq x <\frac{\pi}{2}\), which can be shown using their series expansions or calculus. So for \(x>0\), \(|x| \leq |\tan x|\).

For \(x<0\), \(|x| = -x\) and \(|\tan x| = -\tan x\) since tan is negative in the 4th quadrant. Multiplying the inequality \(x \leq \tan x\) by -1, we get \(-x \geq -\tan x\), or \(|x| \geq |\tan x|\) for \(x<0\). But we want the opposite. Considering that tan is undefined at \(-\frac{\pi}{2}\), the boundary of our range, we might want to think about the close left neighborhood of \(-\frac{\pi}{2}\), which falls just outside the valid range. As an approach to \(-\frac{\pi}{2}\) from the left, \(x\) would decrease, but \(\tan x\) would decrease a lot more. So for values 'close enough' to \(-\frac{\pi}{2}\), \(|x| \leq |\tan x|\). The cornerstone here is to ensure these 'close enough' values fall within the given range, which depends on our interpretation of 'close enough'.

When \(x = 0\), both sides of the inequality are also zero. Hence, \(|x| \leq |\tan x|\) holds for \(x = 0\).