Tan Alpha Beta B9A981

1. **State the problem:** Given $\tan \alpha = \frac{x}{x + 1}$ and $\tan \beta = \frac{1}{2x + 1}$, find the value of $\alpha + \beta$. 2. **Formula used:** Recall the tangent addition formula: $$\tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta}$$ 3. **Substitute the given values:** $$\tan(\alpha + \beta) = \frac{\frac{x}{x+1} + \frac{1}{2x+1}}{1 - \left(\frac{x}{x+1}\right) \left(\frac{1}{2x+1}\right)}$$ 4. **Simplify numerator:** $$\frac{x}{x+1} + \frac{1}{2x+1} = \frac{x(2x+1)}{(x+1)(2x+1)} + \frac{x+1}{(x+1)(2x+1)} = \frac{x(2x+1) + (x+1)}{(x+1)(2x+1)}$$ Calculate numerator inside fraction: $$x(2x+1) + (x+1) = 2x^2 + x + x + 1 = 2x^2 + 2x + 1$$ So numerator is: $$\frac{2x^2 + 2x + 1}{(x+1)(2x+1)}$$ 5. **Simplify denominator:** $$1 - \left(\frac{x}{x+1}\right) \left(\frac{1}{2x+1}\right) = 1 - \frac{x}{(x+1)(2x+1)} = \frac{(x+1)(2x+1) - x}{(x+1)(2x+1)}$$ Calculate numerator inside fraction: $$(x+1)(2x+1) - x = (2x^2 + x + 2x + 1) - x = 2x^2 + 3x + 1 - x = 2x^2 + 2x + 1$$ So denominator is: $$\frac{2x^2 + 2x + 1}{(x+1)(2x+1)}$$ 6. **Combine numerator and denominator:** $$\tan(\alpha + \beta) = \frac{\frac{2x^2 + 2x + 1}{(x+1)(2x+1)}}{\frac{2x^2 + 2x + 1}{(x+1)(2x+1)}}$$ 7. **Cancel common factors:** $$\tan(\alpha + \beta) = \frac{\cancel{\frac{2x^2 + 2x + 1}{(x+1)(2x+1)}}}{\cancel{\frac{2x^2 + 2x + 1}{(x+1)(2x+1)}}} = 1$$ 8. **Interpret result:** Since $\tan(\alpha + \beta) = 1$, then $$\alpha + \beta = \frac{\pi}{4}$$ because tangent of $\frac{\pi}{4}$ is $1$. **Final answer:** $$\boxed{\frac{\pi}{4}}$$