Odpowiedź :
[tex]tg\alpha=\frac{sin\alpha}{cos\alpha}\\sin^2\alpha+cos^2\alpha=1\\[/tex]
1.
[tex]\frac1{1+tg^2\alpha}=\frac1{1+\frac{sin^2\alpha}{cos^2\alpha}}=\frac1{\frac{cos^2\alpha}{cos^2\alpha}+\frac{sin^2\alpha}{cos^2\alpha}}=\frac1{\frac{sin^2\alpha+cos^2\alpha}{cos^2\alpha}}=1*\frac{cos^2\alpha}{sin^2\alpha+cos^2\alpha}=1*\frac{cos^2\alpha}1=cos^2\alpha[/tex]
2.
[tex]\frac1{cos^2\alpha}-1=\frac1{1-sin^2\alpha}-1=\frac{1}{1-sin^2\alpha}-\frac{1-sin^2\alpha}{1-sin^2\alpha}=\frac{1-(1-sin^2\alpha)}{1-sin^2\alpha}=\frac{1-1+sin^2\alpha}{1-sin^2\alpha}=\frac{sin^2\alpha}{1-sin^2\alpha}=\frac{sin^2\alpha}{sin^2\alpha+cos^2\alpha-sin^2\alpha}=\frac{sin^2\alpha}{cos^2\alpha}=(\frac{sin\alpha}{cos\alpha})^2=tg^2\alpha[/tex]