Odpowiedź
[tex]\displaystyle{ tg (\alpha) = \frac {2} {3} }\\\\\displaystyle{ ctg (\alpha) = \frac {1} { \: tg (\alpha) \: } = \frac {3} {2} }[/tex]
[tex]\displaystyle{ sin (\alpha) = \frac {2} { \:\: \displaystyle{ \sqrt{2^2 + 3^2 \,} } \:\: } } = \frac {2} { \:\: \displaystyle{ \sqrt{4 + 9 \, } } \:\: } } = \frac {2} { \:\: \displaystyle{ \sqrt{13 \,} } \:\: } }\\\\\\\displaystyle{ cos (\alpha) = \frac {3} { \:\: \displaystyle{ \sqrt{2^2 + 3^2 \,} } \:\: } } = \frac {3} { \:\: \displaystyle{ \sqrt{4 + 9 \, } } \:\: } } = \frac {3} { \:\: \displaystyle{ \sqrt{13 \,} } \:\: } }[/tex]
[tex]\displaystyle{ sec (\alpha) = \frac {1} { \:\: cos (\alpha) \:\: } = \frac {1} { \:\:\: \displaystyle{ \frac {3} { \: \sqrt{13 \, } \: } } \:\:\: } = \frac { \:\: \sqrt{13 \, } \:\: } {3} } }\\\\\\\displaystyle{ cosec (\alpha) = \frac {1} { \:\: sin (\alpha) \:\: } = \frac {1} { \:\:\: \displaystyle{ \frac {2} { \: \sqrt{13 \, } \: } } \:\:\: } = \frac { \:\: \sqrt{13 \, } \:\: } {2} } }[/tex]
