Given the function $\psi(x) = \tan x$, let's evaluate the expressions step-by-step.

1. Evaluate $\psi\left(\frac{\pi}{6}\right)$: $\psi\left(\frac{\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right)$ Using the known value: $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$ Thus, $\psi\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$

2. Evaluate $\psi\left(x - \frac{\pi}{2}\right)$: $\psi\left(x - \frac{\pi}{2}\right) = \tan\left(x - \frac{\pi}{2}\right)$ Using the tangent identity $\tan\left(x - \frac{\pi}{2}\right) = -\cot x$, we get: $\psi\left(x - \frac{\pi}{2}\right) = -\cot x$

3. Evaluate $\psi(-x)$: $\psi(-x) = \tan(-x)$ Using the odd function property of tangent, $\tan(-x) = -\tan x$, we get: $\psi(-x) = -\tan x$

4. Express $\psi(2x)$ as a function of $\psi(x)$: We want to express $\tan(2x)$ in terms of $\tan(x)$. Using the double-angle identity for tangent: $\tan(2x) = \frac{2\tan x}{1 - \tan^2 x}$ Since $\psi(x) = \tan x$, let $\psi(x) = t$. Then: $\psi(2x) = \frac{2t}{1 - t^2}$

So, summarizing the results:

1. $\psi\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$
2. $\psi\left(x - \frac{\pi}{2}\right) = -\cot x$
3. $\psi(-x) = -\tan x$
4. $\psi(2x) = \frac{2\psi(x)}{1 - (\psi(x))^2}$

Would you like more details on any of these steps or have any questions?

Here are 5 relative questions for further exploration:

1. How would you evaluate $\psi\left(\frac{\pi}{4}\right)$?
2. What is the value of $\psi\left(\frac{\pi}{3}\right)$?
3. How does the function $\psi(x)$ behave as $x$ approaches $\frac{\pi}{2}$?
4. Can you derive the identity for $\psi(3x)$ in terms of $\psi(x)$?
5. What are the critical points of $\psi(x) = \tan x$ within the interval $0 \leq x < \pi$?

Tip: When working with trigonometric functions, knowing and using fundamental identities can significantly simplify the problem-solving process.