Question

Найдите значение выражений!
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Answer 1

Решение:

1)

1 - sin α · cos α · ctg α = 1 - sin α · cos α ·  cos sin α = 1 - cos²α = sin²α

При  sin α =  √3 : 3

sin² α  = 1/3

2)

$\dfrac{tg~\alpha + ctg~\alpha}{tg~\alpha - ctg~\alpha} = \dfrac{tg~\alpha + \dfrac{1}{tg~\alpha} }{tg~\alpha - \dfrac{1}{tg~\alpha} } = \dfrac{tg^2\alpha + 1}{tg^2\alpha - 1}$

При tg α = 2/3

$\dfrac{tg^2\alpha + 1}{tg^2\alpha - 1}= \dfrac{\dfrac{4}{9} +1}{\dfrac{4}{9} -1} = \dfrac{13}{-5} = -2.6$

3)

$\dfrac{sin~\alpha- cos~\alpha}{sin~\alpha+ cos~\alpha} = \dfrac{tg~\alpha - 1}{tg~\alpha + 1}$

При tg α = 2/5 = 0.4

$\dfrac{tg~\alpha - 1}{tg~\alpha + 1} = \dfrac{0.4-1}{0.4+1} = \dfrac{-0.6}{1.4}= -\dfrac{3}{7}$

4)

$\dfrac{sin~\alpha \cdot cos~\alpha}{sin^2 \alpha- cos^2\alpha } = \dfrac{tg~\alpha}{tg^2\alpha - 1}$

При tg α = 3/2 = 1.5

$\dfrac{tg~\alpha}{tg^2\alpha - 1} = \dfrac{1.5}{2.25 - 1} = \dfrac{1.5}{1.25}= 1.2$

5)

$\dfrac{1 - tg^2\alpha}{1 - ctg^2\alpha} =\dfrac{1 - \dfrac{sin^2\alpha}{cos^2\alpha} }{1- \dfrac{cos^2\alpha}{sin^2\alpha} }= \dfrac{sin^2\alpha\cdot (cos^2\alpha- sin^2\alpha)}{cos^2\alpha \cdot(sin^2\alpha - cos^2\alpha)}= \\\\ = \dfrac{sin^2\alpha\cdot (cos^2\alpha- sin^2\alpha)}{-cos^2\alpha \cdot(cos^2\alpha-sin^2\alpha )}= \dfrac{sin^2\alpha}{sin^2\alpha - 1}$

При sin α = 2/3

$\dfrac{sin^2\alpha}{sin^2\alpha - 1} =\dfrac{\dfrac{4}{9} }{\dfrac{4}{9} - 1} =\dfrac{4}{-5} = -0.8$

6)

$\dfrac{1 - tg^2\alpha}{1 - ctg^2\alpha} =\dfrac{1 - \dfrac{sin^2\alpha}{cos^2\alpha} }{1- \dfrac{cos^2\alpha}{sin^2\alpha} }= \dfrac{sin^2\alpha\cdot (cos^2\alpha- sin^2\alpha)}{cos^2\alpha \cdot(sin^2\alpha - cos^2\alpha)}= \\\\ = \dfrac{sin^2\alpha\cdot (cos^2\alpha- sin^2\alpha)}{-cos^2\alpha \cdot(cos^2\alpha-sin^2\alpha )}= \dfrac{1 -cos^2\alpha}{-cos^2\alpha} = \dfrac{cos^2\alpha - 1 }{cos^2\alpha}$

При cos α = -1/3

$\dfrac{cos^2\alpha - 1 }{cos^2\alpha} = \dfrac{\dfrac{1}{9}-1 }{\dfrac{1}{9} } = -8$