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Решите задачу:
(Sinα-tgα)·(cosα-1)⁻¹=tgα; (sinα-tgα)·(cosα-1)⁻¹=(sinα-sinα/cosα)·1/(cosα-1)= =(sinα·cosα-sinα)/[cosα(cosα-1)]=[sinα(cosα-1)]/[cosα(cosα-1)]= =sinα/cosα=tgα