3.4.2 Hence, or otherwise determine the maximum value of: \( \frac{(1+\tan \beta)^{2}}{1+\tan ^{2} \beta} \)

To determine the maximum value of \( \frac{(1+\tan \beta)^{2}}{1+\tan ^{2} \beta} \), we can start by simplifying the expression. Let \( x = \tan \beta \). Then our expression transforms to: \[ f(x) = \frac{(1+x)^{2}}{1+x^{2}} \] Now we simplify it: \[ f(x) = \frac{1 + 2x + x^{2}}{1 + x^{2}} \] We can split this into parts: \[ f(x) = \frac{1 + x^{2}}{1 + x^{2}} + \frac{2x}{1 + x^{2}} = 1 + \frac{2x}{1 + x^{2}} \] Next, we need to maximize \( \frac{2x}{1 + x^{2}} \). Let: \[ g(x) = \frac{2x}{1 + x^{2}} \] We can differentiate \( g(x) \) to find the critical points: \[ g'(x) = \frac{(1 + x^{2})(2) - 2x(2x)}{(1 + x^{2})^2} = \frac{2(1 + x^{2} - 2x^{2})}{(1 + x^{2})^2} = \frac{2(1 - x^{2})}{(1 + x^{2})^2} \] Setting \( g'(x) = 0 \): \[ 1 - x^{2} = 0 \implies x^{2} = 1 \implies x = 1 \, \text{or} \, x = -1 \] Now, we can evaluate \( g(x) \) at these points: For \( x = 1 \): \[ g(1) = \frac{2(1)}{1 + 1^{2}} = \frac{2}{2} = 1 \] For \( x = -1 \): \[ g(-1) = \frac{2(-1)}{1 + (-1)^{2}} = \frac{-2}{2} = -1 \] Next, let's check the limits as \( x \) approaches \( \pm \infty \): \[ \lim_{x \to +\infty} g(x) = 0 \quad \text{and} \quad \lim_{x \to -\infty} g(x) = 0 \] Thus, the maximum value \( g(x) \) achieves occurs at \( x = 1 \): \[ g(1) = 1 \] Finally, adding this to the constant part of \( f(x) \): \[ f(1) = 1 + 1 = 2 \] Therefore, the maximum value of the original expression \( \frac{(1+\tan \beta)^{2}}{1+\tan ^{2} \beta} \) is: \[ \boxed{2} \]