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Данякеп
@Данякеп
September 2021
1
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Всё на фото..............
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oganesbagoyan
Verified answer
Task/26548423
---------------------
10.
tg(α+5π/2) = tg(5π/2+α) =tg
(
2π +(π/2+α
)
)
=tg(
π/2+α) = -ctgα = -1/tgα =|| tg
α=0,4 || = -1/(0,4)
= - 2,5 .
---------------
11.
5sin²α +13cos²α =6 ⇔ 5sin²α +13cos²α =6(sin²α +cos²α) ⇔sin²α = 12cos²α⇒
sin²α / cos²α = 12 ⇔
tg²α =12.
---------------
12.
(3cos
α - 4sinα)
/
(2sinα -5cosα)
=
* * *
Pаз определена функция tgα ( tgα =3), значить
cosα ≠ 0
числитель и знаменатель дроби разделим на cosα ≠ 0 * * *
(3cosα/ cosα - 4sinα/cosα) /
(2sinα/cos
α
-5cosα/ cos
α
) =
(3
- 4tgα)
/ (2tg
α -5) =(3 - 4*3) / (2*3 -5) = -9 /1
= - 9.
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Answers & Comments
Verified answer
Task/26548423---------------------
10.
tg(α+5π/2) = tg(5π/2+α) =tg(2π +(π/2+α)) =tg(π/2+α) = -ctgα = -1/tgα =|| tgα=0,4 || = -1/(0,4) = - 2,5 .
---------------
11.
5sin²α +13cos²α =6 ⇔ 5sin²α +13cos²α =6(sin²α +cos²α) ⇔sin²α = 12cos²α⇒
sin²α / cos²α = 12 ⇔ tg²α =12.
---------------
12.
(3cosα - 4sinα) / (2sinα -5cosα) =
* * *
Pаз определена функция tgα ( tgα =3), значить cosα ≠ 0
числитель и знаменатель дроби разделим на cosα ≠ 0 * * *
(3cosα/ cosα - 4sinα/cosα) / (2sinα/cosα -5cosα/ cosα) =
(3 - 4tgα) / (2tgα -5) =(3 - 4*3) / (2*3 -5) = -9 /1 = - 9.