1. The problem states the identity for the tangent of a sum of two angles: $$\tan(x + y) = s$$ and asks to verify or understand the formula involving $$\tan a$$ and $$\tan y$$. 2. The formula given is: $$\frac{\tan x + \tan y}{1 - \tan x \cdot \tan y} = \tan(x + y)$$ 3. This is a standard trigonometric identity for the tangent of the sum of two angles. It means that the tangent of the sum of angles $$x$$ and $$y$$ equals the fraction with numerator $$\tan x + \tan y$$ and denominator $$1 - \tan x \tan y$$. 4. To understand why this formula works, recall that tangent is sine over cosine: $$\tan \theta = \frac{\sin \theta}{\cos \theta}$$ 5. Using the sine and cosine addition formulas: $$\sin(x + y) = \sin x \cos y + \cos x \sin y$$ $$\cos(x + y) = \cos x \cos y - \sin x \sin y$$ 6. Dividing $$\sin(x + y)$$ by $$\cos(x + y)$$ gives: $$\tan(x + y) = \frac{\sin x \cos y + \cos x \sin y}{\cos x \cos y - \sin x \sin y}$$ 7. Dividing numerator and denominator by $$\cos x \cos y$$: $$\tan(x + y) = \frac{\frac{\sin x}{\cos x} + \frac{\sin y}{\cos y}}{1 - \frac{\sin x}{\cos x} \cdot \frac{\sin y}{\cos y}} = \frac{\tan x + \tan y}{1 - \tan x \tan y}$$ 8. Therefore, the formula is correct and shows how to compute $$\tan(x + y)$$ using $$\tan x$$ and $$\tan y$$. Final answer: $$\tan(x + y) = \frac{\tan x + \tan y}{1 - \tan x \tan y} = s$$