Question

Добрый день Нужна ваша
Докажите тождество: = $\frac{ \tan \alpha + \tan \beta }{1 - (\tan \alpha \times \tan \beta) }$
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$\tan( \alpha + \beta )$

Answer 1

$\displaystyle\\\tan(\alpha+\beta) = \frac{\sin(\alpha+\beta)}{\cos(\alpha+\beta)} = \frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\alpha\cos\beta-\sin\alpha\sin\beta}$

Разделим все на $\cos\alpha\cos\beta$

$\displaystyle\\\tan(\alpha+\beta) = \frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

Answer 2

tg(α+β)=sin(α+β)/cos(α+β)=(sinα*cosβ+cosα*sinβ)/(cosα*cosβ-sinα*sinβ)=

(sinα*cosβ/(cosα*cosβ)+(cosα*sinβ)/(cosα*cosβ))/(cosα*cosβ/(cosα*cosβ)-sinα*sinβ/(cosα*cosβ))=(tgα+tgβ)/(1-(tgα*+tgβ)) Доказано.