Given the conditions \[\alpha + \beta = \frac{\pi}{2}, \quad \beta + \gamma = \alpha.\] From the first condition, we derive \[\alpha = \frac{\pi}{2} - \beta.\] Substituting this into the second condition yields \[\beta + \gamma = \frac{\pi}{2} - \beta.\] Solving for \(\gamma\) gives \[\gamma = \frac{\pi}{2} - 2\beta.\] We aim to find \(\tan\alpha\). Using \(\alpha = \frac{\pi}{2} - \beta\), we apply the tangent identity: \[\tan\alpha = \tan\left(\frac{\pi}{2} - \beta\right) = \cot\beta.\] Next, we express \(\cot\beta\) in terms of \(\tan\beta\): \[\cot\beta = \frac{1}{\tan\beta}.\] From \(\gamma = \frac{\pi}{2} - 2\beta\), we have \[\tan\gamma = \tan\left(\frac{\pi}{2} - 2\beta\right) = \cot(2\beta).\] Applying the double angle identity for cotangent: \[\cot(2\beta) = \frac{1 - \tan^2\beta}{2\tan\beta}.\] Substituting this into the equation for \(\tan\alpha\): \[\tan\alpha = \tan\beta + 2\tan\gamma = \tan\beta + 2 \cdot \frac{1 - \tan^2\beta}{2\tan\beta}.\] Simplifying the expression: \[\tan\alpha = \tan\beta + \frac{1 - \tan^2\beta}{\tan\beta}.\] The value of \(\tan\alpha\) is therefore \[\tan\alpha = \tan\beta + 2\tan\gamma.\]