Answer :
[tex]sin^2 \alpha +cos^2 \alpha =1\ \ \ and\ \ \ tan \alpha = \frac{\big{sin \alpha }}{\big{cos \alpha }} \\\\sin \alpha = \frac{8}{9}\\\\ \Rightarrow\ \ \ (\frac{8}{9})^2+cos^2 \alpha =1\ \ \ \Rightarrow\ \ \ cos^2 \alpha =1- \frac{64}{81} \ \ \ \Rightarrow\ \ \ cos^2 \alpha = \frac{17}{81} \\\\ \alpha \ \in\ (0^0;90^0)\ \ \ \Rightarrow\ \ \ cos \alpha >0\ \ \ \Rightarrow\ \ \ cos \alpha = \frac{ \sqrt{17} }{9} \\\\[/tex]
[tex]tan \alpha = \frac{8}{9}: \frac{ \sqrt{17} }{9} =\frac{8}{9}\cdot \frac{9}{\sqrt{17}}= \frac{8}{\sqrt{17}}= \frac{8\cdot \sqrt{17}}{\sqrt{17}\cdot \sqrt{17}}=\frac{8\cdot \sqrt{17}}{17}[/tex]
[tex]tan \alpha = \frac{8}{9}: \frac{ \sqrt{17} }{9} =\frac{8}{9}\cdot \frac{9}{\sqrt{17}}= \frac{8}{\sqrt{17}}= \frac{8\cdot \sqrt{17}}{\sqrt{17}\cdot \sqrt{17}}=\frac{8\cdot \sqrt{17}}{17}[/tex]
