If tan alpha = frac{x^2 - x}{x^2 - x + 1} and | vedclass

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(A) Let $t = x^2 - x$. Then $	an \alpha = \frac{t}{t + 1}$ and $	an \beta = \frac{1}{2t + 1}$.Using the formula $	an(\alpha + \beta) = \frac{	an \

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(A) Let $t = x^2 - x$. Then $	an \alpha = \frac{t}{t + 1}$ and $	an \beta = \frac{1}{2t + 1}$.Using the formula $	an(\alpha + \beta) = \frac{	an \alpha + 	an \beta}{1 - 	an \alpha 	an \beta}$,we substitute the values:$	an(\alpha + \beta) = \frac{\frac{t}{t + 1} + \frac{1}{2t + 1}}{1 - \left(\frac{t}{t + 1}ight) \left(\frac{1}{2t + 1}ight)}$$= \frac{\frac{t(2t + 1) + (t + 1)}{(t + 1)(2t + 1)}}{\frac{(t + 1)(2t + 1) - t}{(t + 1)(2t + 1)}}$$= \frac{2t^2 + t + t + 1}{2t^2 + 2t + t + 1 - t}$$= \frac{2t^2 + 2t + 1}{2t^2 + 2t + 1} = 1$Therefore,$	an(\alpha + \beta) = 1$.

If $\tan \alpha = \frac{x^2 - x}{x^2 - x + 1}$ and $\tan \beta = \frac{1}{2x^2 - 2x + 1}$ $(x \ne 0, 1)$,where $0 < \alpha, \beta < \frac{\pi}{2}$,then $\tan(\alpha + \beta)$ has the value equal to:

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