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lark1972
@lark1972
July 2022
2
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Помогите ! Нужно сделать а и б!
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oganesbagoyan
Verified answer
а)
= tq²α*sin²α/(tq²α -sin²α) =cos²α*tq²α*sin²α/sin²α(1 -cos²α)=sin²α*sin²α/sin²α*sin²α
=1.
------
б)
=( tqα/(1-tq²α)) * ((ctq²α -1)/ctqα )= (1/2)*tq2α* 2ctq2α
=1.
2 votes
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Evklid61
Verified answer
Первое не твое, не нужно переписывать
1 votes
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Answers & Comments
Verified answer
а) = tq²α*sin²α/(tq²α -sin²α) =cos²α*tq²α*sin²α/sin²α(1 -cos²α)=sin²α*sin²α/sin²α*sin²α =1.------
б) =( tqα/(1-tq²α)) * ((ctq²α -1)/ctqα )= (1/2)*tq2α* 2ctq2α =1.
Verified answer
Первое не твое, не нужно переписывать