zad.3
\sin\alpha=\frac{12}{13}
\sin^2\alpha+ \cos^2\alpha=1
\cos^2\alpha=1-sin^2\alpha=1-(\frac{12}{13})^2=1-\frac{144}{169}=\frac{25}{169}
\cos\alpha=\sqrt{\frac{25}{169}}=\frac{5}{13} lub \cos\alpha=-\sqrt{\frac{25}{169}}=-\frac{5}{13}
\tan\alpha=\frac{\sin\alpha}{\cos\alpha}
\tan\alpha=\frac{\frac{12}{13}}{\frac{5}{13}}=2,4 lub \tan\alpha=\frac{\frac{12}{13}}{-\frac{5}{13}}=-2,4
\cot\alpha=\frac{1}{\tan\alpha}
\cot\alpha=\frac{1}{2,4}=\frac{1}{\frac{24}{10}}=\frac{1}{\frac{12}{5}}=\frac{5}{12} lub \cot\alpha=\frac{1}{-2,4}=\frac{1}{-\frac{24}{10}}=\frac{1}{-\frac{12}{5}}=-\frac{5}{12}