▬ 1-cos²α▬ cos²α/(1-sin²α)▬ (1+tgα)²+(1-tgα)²▬ (1-sin²α/1-cos²α)+tgα+ctgα▬ tgα/(1+tg²α)

Question

▬ 1-cos²α
▬ cos²α/(1-sin²α)
▬ (1+tgα)²+(1-tgα)²
▬ (1-sin²α/1-cos²α)+tgα+ctgα
▬ tgα/(1+tg²α)

Answer 1

Решение во вложения.......

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