Докажете тъждествата

[tedkoos]

1. tg^2 α+ cotg^2 α= 1/sin^2 α + 1/cos^2 α - 2
2.sin^6 α +cos^6 α = 1-3*sin^2 α * cosα

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Задача 1.
[tex]tg^2\alpha+cotg^2\alpha=\frac{1}{sin^2\alpha}+\frac{1}{cos^2\alpha}-2[/tex]
[tex]\frac{sin^2\alpha}{cos^2\alpha}+\frac{cos^2\alpha}{sin^2\alpha}=\frac{cos^2\alpha+sin^2\alpha}{sin^2\alpha cos^2\alpha}-2[/tex]
[tex]\frac{sin^4\alpha+cos^4\alpha}{sin^2\alpha cos^2\alpha}=\frac{1}{sin^2\alpha cos^2\alpha}-2[/tex]

[tex]\frac{(sin^2\alpha)^2+(cos^2\alpha)^2+2sin^2\alpha cos^2\alpha-2sin^2\alpha cos^2\alpha}{sin^2\alpha cos^2\alpha}=\frac{1-2sin^2\alpha cos^2\alpha}{sin^2\alpha cos^2\alpha}[/tex]

[tex]\frac{(sin^2\alpha+cos^2\alpha)^2-2sin^2\alpha cos^2\alpha}{sin^2\alpha cos^2\alpha}=\frac{1-2sin^2\alpha cos^2\alpha}{sin^2\alpha cos^2\alpha}[/tex]

[tex]\frac{1-2sin^2\alpha cos^2\alpha}{sin^2\alpha cos^2\alpha}=\frac{1-2sin^2\alpha cos^2\alpha}{sin^2\alpha cos^2\alpha}[/tex]

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Задача 2.
[tex]sin^6\alpha+cos^6\alpha=1-3sin^2\alpha cos^2\alpha[/tex]
[tex](sin^2\alpha)^3+(cos^2\alpha)^3=1-3sin^2\alpha cos^2\alpha[/tex]
[tex](sin^2\alpha+cos^2\alpha)((sin^2\alpha)^2-sin^2\alpha cos^2\alpha+(cos^2\alpha)^2)=1-3sin^2\alpha cos^2\alpha[/tex]
[tex](sin^2\alpha)^2-sin^2\alpha cos^2\alpha+(cos^2\alpha)^2+2sin^2\alpha cos^2\alpha-2sin^2\alpha cos^2\alpha=1-3sin^2\alpha cos^2\alpha[/tex]
[tex](sin^2\alpha+cos^2\alpha)^2-3sin^2\alpha cos^2\alpha=1-3sin^2\alpha cos^2\alpha[/tex]
[tex]1-3sin^2\alpha cos^2\alpha=1-3sin^2\alpha cos^2\alpha[/tex]