Funkcje trygonometryczne i chyba.... tożsamość

[Lorek]

Podnieś do kwadratu to co masz dane

[natkoza]

\(\displaystyle{ tg^{2}\alpha+ctg^{2}\alpha=\frac{sin^{2}\alpha}{cos^{2}\alpha}+\frac{cos^{2}\alpha}{sin^{2}\alpha}=\frac{sin^{4}\alpha+cos^{4}\alpha}{sin^{2}\alpha cos^{2}\alpha}=\frac{(sin^{2}\alpha+cos^{2}\alpha)^{2}-2sin^{2}\alpha cos^{2}\alpha}{ (sin\alpha cos\alpha)^{2}}=\frac{1-2(sin\alpha cos\alpha)^{2}}{(sin\alpha cos\alpha)^{2}}=\frac{1-2(\frac{1}{4})^{2}}{\frac{1}{4})^{2}}=\frac{1-\frac{1}{8}}{\frac{1}{16}}=\frac{\frac{7}{8}}{\frac{1}{16}}=\frac{7}{8}\cdot 16=14 \\
tg\alpha+ctg\alpha=4 \frac{sin\alpha}{cos\alpha}+\frac{cos\alpha}{sin\alpha}=4 \frac{sin^{2}\alpha+cos^{2}\slpha}{sin\alpha cos\alpha}=4 \frac{1}{sin\alpha cos\alpha}=4 sin\alpha cos\alpha=\frac{1}{4}}\)
lub szybciej:
\(\displaystyle{ (tg^{2}\alpha +ctg^{2}\alpha)=(tg\alpha+ctg\alpha)^{2}-2tg\alpha cgt\alpha=16-2=14}\)

[soku11]

Jesli \(\displaystyle{ \alpha\in(0;\frac{\pi}{2})}\) to wtedy \(\displaystyle{ sin\alpha,\ cos\alpha,\ tan\alpha,\ cot\alpha}\) jest zawsze dodatni POZDRO