Paso 1:
Si $\tan \alpha = \frac{1}{\sqrt{x(x^2 + x + 1)}}$, $\tan \beta = \frac{\sqrt{x}}{\sqrt{x^2 + x + 1}}$ y $\tan \gamma = \frac{\sqrt{x^2 + x + 1}}{x\sqrt{x}}$, demuestre que $\alpha + \beta = \gamma$.