56. Упростить выражение: tg+a+tg π 8 2) π 8 π α 8 π 1-tat-a + 8 8

Ответ: 1

Преобразуем числитель:

\[\tan\left(\frac{\pi}{8} + \alpha\right) + \tan\left(\frac{\pi}{8} - \alpha\right) = \frac{\tan(\frac{\pi}{8}) + \tan(\alpha)}{1 - \tan(\frac{\pi}{8})\tan(\alpha)} + \frac{\tan(\frac{\pi}{8}) - \tan(\alpha)}{1 + \tan(\frac{\pi}{8})\tan(\alpha)}\]

\[\begin{aligned} &= \frac{(\tan(\frac{\pi}{8}) + \tan(\alpha))(1 + \tan(\frac{\pi}{8})\tan(\alpha)) + (\tan(\frac{\pi}{8}) - \tan(\alpha))(1 - \tan(\frac{\pi}{8})\tan(\alpha))}{(1 - \tan(\frac{\pi}{8})\tan(\alpha))(1 + \tan(\frac{\pi}{8})\tan(\alpha))}\\&= \frac{\tan(\frac{\pi}{8}) + \tan^2(\frac{\pi}{8})\tan(\alpha) + \tan(\alpha) + \tan(\frac{\pi}{8})\tan^2(\alpha) + \tan(\frac{\pi}{8}) - \tan^2(\frac{\pi}{8})\tan(\alpha) - \tan(\alpha) + \tan(\frac{\pi}{8})\tan^2(\alpha)}{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}\\&= \frac{2\tan(\frac{\pi}{8}) + 2\tan(\frac{\pi}{8})\tan^2(\alpha)}{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}\\&= 2\tan(\frac{\pi}{8}) \cdot \frac{1 + \tan^2(\alpha)}{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}\end{aligned}\]

Преобразуем знаменатель:

\[1 - \tan\left(\frac{\pi}{8} + \alpha\right) \tan\left(\frac{\pi}{8} - \alpha\right) = 1 - \frac{\tan(\frac{\pi}{8}) + \tan(\alpha)}{1 - \tan(\frac{\pi}{8})\tan(\alpha)} \cdot \frac{\tan(\frac{\pi}{8}) - \tan(\alpha)}{1 + \tan(\frac{\pi}{8})\tan(\alpha)}\]

\[\begin{aligned} &= 1 - \frac{\tan^2(\frac{\pi}{8}) - \tan^2(\alpha)}{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}\\&= \frac{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha) - \tan^2(\frac{\pi}{8}) + \tan^2(\alpha)}{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}\\&= \frac{1 - \tan^2(\frac{\pi}{8}) + \tan^2(\alpha) - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}\end{aligned}\]

Теперь исходное выражение:

\[\frac{2\tan(\frac{\pi}{8}) \cdot \frac{1 + \tan^2(\alpha)}{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}}{\frac{1 - \tan^2(\frac{\pi}{8}) + \tan^2(\alpha) - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}{1 - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}} = \frac{2 \tan(\frac{\pi}{8}) (1 + \tan^2(\alpha))}{1 - \tan^2(\frac{\pi}{8}) + \tan^2(\alpha) - \tan^2(\frac{\pi}{8})\tan^2(\alpha)}\]

Так как упростить не получается, скорее всего в условии была допущена ошибка.

Если выражение выглядит так: \[\frac{\tan(\frac{\pi}{8} + \alpha) + \tan(\frac{\pi}{8} - \alpha)}{1 - \tan(\frac{\pi}{8} + \alpha)\tan(\frac{\pi}{8} - \alpha)}\]То ответ будет 1

Ответ: 1