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наташа29242001
@наташа29242001
August 2022
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Пожалуйста с 22 по 25 пример .
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oganesbagoyan
Verified answer
22.
(1-cosα)/(1+cosα) -(1-2cosα)/sin²α=
(1-cosα)²/(1+cosα)(1-cosα) -(1-2cosα)/sin²α=
(1-cosα)²/(1-cos²α) -(1-2cosα)/sin²α =
(1-2cosα+cos²α)/sin²α -
(1-2cosα)/sin²α =
(1-2cosα+cos²α -1+2cosα)/sin²α =cos²α/sin²α =ctq²α. →А)
-------
23.
tq²α - sin²α - tq²α*sin²α=tq²α-tq²α*sin²α-sin²α =tq²α(1-sin²α )
- sin²α =
tq²α*cos²α - sin²α =sin²α -sin²α =0 .
→Е)
-------
24. Можно корректировать пример ?
(sin(α+β) -cosα*sinβ) /(sin(α-β) +cosα*sinβ) =
(sinα*cosβ+cosα*sinβ -cosα*sinβ) /(sinα*cosβ-cosα*sinβ +cosα*sinβ) =
sinα*cosβ/sinα*cosβ =1. →А)
-------
25.
(sin
α+cosα)² -sin2α=sin²α+2sinα*cosα+cos²α -sin2α=
(sin²α+cos²α) +2sinα*cosα -sin2α =1 +sin2α-sin2α =1. →С)
2 votes
Thanks 1
oganesbagoyan
Буду рад быть полезным !
наташа29242001
спасибо огромное,очень выручили!)))
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Answers & Comments
Verified answer
22.(1-cosα)/(1+cosα) -(1-2cosα)/sin²α=
(1-cosα)²/(1+cosα)(1-cosα) -(1-2cosα)/sin²α=
(1-cosα)²/(1-cos²α) -(1-2cosα)/sin²α =
(1-2cosα+cos²α)/sin²α -(1-2cosα)/sin²α =
(1-2cosα+cos²α -1+2cosα)/sin²α =cos²α/sin²α =ctq²α. →А)
-------
23.
tq²α - sin²α - tq²α*sin²α=tq²α-tq²α*sin²α-sin²α =tq²α(1-sin²α ) - sin²α =
tq²α*cos²α - sin²α =sin²α -sin²α =0 . →Е)
-------
24. Можно корректировать пример ?
(sin(α+β) -cosα*sinβ) /(sin(α-β) +cosα*sinβ) =
(sinα*cosβ+cosα*sinβ -cosα*sinβ) /(sinα*cosβ-cosα*sinβ +cosα*sinβ) =
sinα*cosβ/sinα*cosβ =1. →А)
-------
25.
(sinα+cosα)² -sin2α=sin²α+2sinα*cosα+cos²α -sin2α=
(sin²α+cos²α) +2sinα*cosα -sin2α =1 +sin2α-sin2α =1. →С)