Skorzystamy z tożsamości trygonometrycznych:
[tex]\sin^2\alpha+\cos^2\alpha=1\,,\qquad \text{tg\,}\alpha=\dfrac{\sin\alpha}{\cos\alpha}\,,\qquad \text{ctg\,}\alpha=\dfrac{1}{\text{tg\,}\alpha}[/tex]
a)
[tex]\sin\alpha=\dfrac14\\\\\left(\frac14\right)^2+\cos^2\alpha=1\\\\\cos^2\alpha=1-\frac{1}{16}\\\\\cos^2\alpha=\frac{15}{16}\qquad\wedge\quad\alpha\in(0^o\,,\ 90^o)\\\\\cos\alpha=\frac{\sqrt{15}}{4}\\\\\\\text{tg\,}\alpha=\dfrac{\frac{1}{4}}{\frac{\sqrt{15}}{4}}=\frac14\cdot\frac4{\sqrt{15}}=\frac1{\sqrt{15}}=\frac{\sqrt{15}}{15} \\\\\\ \text{ctg\,}\alpha=\dfrac{1}{\frac1{\sqrt{15}}}=\sqrt{15}[/tex]
b)
[tex]\text{tg\,}\alpha=\dfrac{2\sqrt2}3\\\\\dfrac{\sin\alpha}{\cos\alpha}= \dfrac{2\sqrt2}3\\\\\sin\alpha=\frac{2\sqrt2}3\cos\alpha\\\\ \left(\frac{2\sqrt2}3\cos\alpha\right)^2+\cos^2\alpha=1\\\\ \frac89\cos^2\alpha+\cos^2\alpha=1\\\\\frac{17}9\cos^2\alpha=1 \qquad/\cdot\frac9{17}\\\\\cos^2\alpha=\frac{9}{17}\qquad\wedge\quad\alpha\in(0^o\,,\ 90^o) \\\\\cos\alpha=\frac3{\sqrt{17}}=\frac{3\sqrt{17}}{17}\\\\\\ \text{ctg\,}\alpha=\dfrac1{\text{ctg\,}\alpha}= \dfrac1{\frac{2\sqrt2}3}=\frac3{2\sqrt2}=\frac{3\sqrt2}4[/tex]