7.
[tex](1+2sin\alpha cos\alpha)(tg\alpha-1)=(sin^2\alpha-cos^2\alpha)(tg\alpha+1)[/tex]
[tex]L=(1+2sin\alpha cos\alpha)(tg\alpha-1)=[/tex]
[tex]tg\alpha-1+2sin\alpha cos\alpha tg\alpha-2sin\alpha cos\alpha=[/tex]
[tex]tg\alpha-1+2sin\alpha cos\alpha\cdot\frac{sin\alpha}{cos\alpha}-2sin\alpha cos\alpha=[/tex]
[tex]tg\alpha-1+2sin^2\alpha-2sin\alpha cos\alpha=[/tex]
[tex]tg\alpha-1+2(1-cos^2\alpha)-2sin\alpha cos\alpha=[/tex]
[tex]tg\alpha-1+2-2cos^2\alpha-2sin\alpha cos\alpha=[/tex]
[tex]tg\alpha+1-2cos^2\alpha-2sin\alpha cos\alpha[/tex]
[tex]P=(sin^2\alpha-cos^2\alpha)(tg\alpha+1)=[/tex]
[tex](1-cos^2\alpha-cos^2\alpha)(tg\alpha+1)=[/tex]
[tex](1-2cos^2\alpha)(tg\alpha+1)=[/tex]
[tex]tg\alpha+1-2cos^2\alpha tg\alpha-2cos^2\alpha=[/tex]
[tex]tg\alpha+1-2cos^2\alpha \cdot\frac{sin\alpha}{cos\alpha}-2cos^2\alpha=[/tex]
[tex]tg\alpha+1-2cos\alpha sin\alpha-2cos^2\alpha[/tex]
[tex]L=P[/tex]
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8.
[tex]\alpha=1918^o=5\cdot360^o+118^o[/tex]
[tex]\alpha=1918^o=10\cdot180^o+118^o[/tex]
[tex]118^o[/tex] - kąt II ćwiartki
[tex]sin\alpha=sin 118^o>0[/tex]
[tex]cos\alpha=cos 118^o<0[/tex]
[tex]tg\alpha=tg 118^o<0[/tex]
[tex]ctg\alpha=ctg 118^o<0[/tex]
Prawdziwa jest A.
