1. State the problem: Given $$y = \frac{1 + \tan x}{1 - \tan x}$$, find $$\frac{dy}{dx}$$ and identify it from the options. 2. Recognize that $$y = \frac{1 + \tan x}{1 - \tan x}$$ can be rewritten using the tangent addition formula: $$\tan(a+b) = \frac{\tan a + \tan b}{1 - \tan a \tan b}$$ Comparing, if we let $$a = x$$ and $$b = 45^\circ$$, since $$\tan 45^\circ = 1$$, then $$y = \tan(x + 45^\circ)$$. 3. Differentiate $$y = \tan(x + 45^\circ)$$ using the chain rule: $$\frac{dy}{dx} = \sec^2(x + 45^\circ) \cdot \frac{d}{dx}(x + 45^\circ) = \sec^2(x + 45^\circ)$$ (Since $$\frac{d}{dx}(x + 45^\circ) = 1$$) 4. Conclusion: $$\frac{dy}{dx} = \sec^2(x + 45^\circ)$$ which corresponds to option (b). Final answer: (b) $$\sec^2(x + 45^\circ)$$.