Introduction:
To prove that the function f(x) = tan(x) is strictly increasing on the interval (-π/2, π/2), we need to show that for any two values a and b in this interval such that a < b,="" f(a)="" />< />

Proof:
Let's consider two arbitrary values a and b in the interval (-π/2, π/2) such that a < />

Mean Value Theorem:
By the Mean Value Theorem, there exists a c in the interval (a, b) such that:
f(b) - f(a) = f'(c)(b - a)

Derivative of tan(x):
The derivative of tan(x) can be found by differentiating it with respect to x. Using the quotient rule, we have:
f'(x) = d/dx(tan(x)) = sec^2(x)

Secant Function:
The secant function, sec(x), is always positive on the interval (-π/2, π/2) since it is the reciprocal of the cosine function, and cosine is positive in this interval.

Proof by Contradiction:
Assume that f(a) ≥ f(b), which means that tan(a) ≥ tan(b). This implies that sec^2(a) ≥ sec^2(b).

Case 1: sec^2(a) > sec^2(b):
If sec^2(a) > sec^2(b), then we can choose a value c in the interval (a, b) such that sec^2(c) is greater than both sec^2(a) and sec^2(b). However, this contradicts the Mean Value Theorem since f'(c) = sec^2(c) would be greater than the value obtained from f'(a) and f'(b) in the equation f(b) - f(a) = f'(c)(b - a).

Case 2: sec^2(a) = sec^2(b):
If sec^2(a) = sec^2(b), then the function f(x) = tan(x) would not be strictly increasing on the interval (-π/2, π/2) as there would be two distinct values a and b such that f(a) = f(b).

Conclusion:
In both cases, our assumption that f(a) ≥ f(b) leads to a contradiction. Therefore, f(x) = tan(x) is strictly increasing on the interval (-π/2, π/2).

Summary:
In summary, we proved that the function f(x) = tan(x) is strictly increasing on the interval (-π/2, π/2) by using the Mean Value Theorem and the properties of the secant function. We showed that assuming f(a) ≥ f(b) leads to a contradiction, concluding that f(x) = tan(x) is strictly increasing on the given interval.