1. **State the problem:** We are given the function $$f(x) = (\tan x)^{2020} + (\tan x)^{2022}$$ defined on the interval $$I = \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[.$$ We want to analyze or understand this function. 2. **Rewrite the function:** Notice that both terms have powers of $$\tan x$$. We can factor out the smaller power: $$f(x) = (\tan x)^{2020} + (\tan x)^{2022} = (\tan x)^{2020} \left(1 + (\tan x)^2\right).$$ 3. **Recall a trigonometric identity:** We know that $$1 + \tan^2 x = \sec^2 x$$. 4. **Substitute the identity:** $$f(x) = (\tan x)^{2020} \cdot \sec^2 x.$$ 5. **Domain considerations:** The function is defined on $$\left]-\frac{\pi}{2}, \frac{\pi}{2}\right[,$$ where $$\tan x$$ and $$\sec x$$ are defined and continuous. 6. **Behavior:** Since $$2020$$ is even, $$(\tan x)^{2020} \geq 0$$ for all $$x$$ in the domain. Also, $$\sec^2 x > 0$$ in the domain. Therefore, $$f(x) \geq 0$$ for all $$x$$ in $$I$$. 7. **Summary:** The function is $$f(x) = (\tan x)^{2020} \sec^2 x$$ on $$I = \left]-\frac{\pi}{2}, \frac{\pi}{2}\right[.$$ It is always non-negative and grows large near the boundaries where $$\tan x$$ and $$\sec x$$ tend to infinity. **Final answer:** $$f(x) = (\tan x)^{2020} \sec^2 x$$ on $$\left]-\frac{\pi}{2}, \frac{\pi}{2}\right[.$$