Question

Помогите пожалуйста!!! ДОКАЖИТЕ ТОЖДЕСТВО.​

Answer 1

Ответ:

$\dfrac{3}{2}=\dfrac{3}{2}$

Объяснение:

$\dfrac{1-sin^{6}\alpha-cos^{6}\alpha}{1-sin^{4}\alpha-cos^{4}\alpha}=\dfrac{3}{2};$

$\dfrac{1-(sin^{6}\alpha+cos^{6}\alpha)}{1-(sin^{4}\alpha+cos^{4}\alpha)}=\dfrac{3}{2};$

$\dfrac{1-((sin^{2}\alpha)^{3}+(cos^{2}\alpha)^{3})}{1-((sin^{2}\alpha)^{2}+(cos^{2}\alpha)^{2})}=\dfrac{3}{2};$

$\dfrac{1-(sin^{2}\alpha+cos^{2}\alpha)((sin^{2}\alpha)^{2}-sin^{2}\alpha \cdot cos^{2}\alpha+(cos^{2}\alpha)^{2})}{1-((sin^{2}\alpha)^{2}+2sin^{2}\alpha \cdot cos^{2}\alpha+(cos^{2}\alpha)^{2}-2sin^{2}\alpha \cdot cos^{2}\alpha)}=\dfrac{3}{2};$

$\dfrac{1-((sin^{2}\alpha)^{2}+2sin^{2}\alpha \cdot cos^{2}\alpha+(cos^{2}\alpha)^{2}-3sin^{2}\alpha \cdot cos^{2}\alpha)}{1-((sin^{2}\alpha+cos^{2}\alpha)^{2}-2sin^{2}\alpha \cdot cos^{2}\alpha)}=\dfrac{3}{2};$

$\dfrac{1-((sin^{2}\alpha+cos^{2}\alpha)^{2}-3sin^{2}\alpha\cdot cos^{2}\alpha)}{1-(1^{2}-2sin^{2}\alpha\cdot cos^{2}\alpha)}=\dfrac{3}{2};$

$\dfrac{1-(1^{2}-3sin^{2}\alpha\cdot cos^{2}\alpha)}{1-1+2sin^{2}\alpha\cdot cos^{2}\alpha}=\dfrac{3}{2};$

$\dfrac{1-1+3sin^{2}\alpha\cdot cos^{2}\alpha}{2sin^{2}\alpha\cdot cos^{2}\alpha}=\dfrac{3}{2};$

$\dfrac{3sin^{2}\alpha\cdot cos^{2}\alpha}{2sin^{2}\alpha\cdot cos^{2}\alpha}=\dfrac{3}{2};$

$\dfrac{3}{2}=\dfrac{3}{2};$

Answer 2

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