Answer:
[tex]\tan \theta = \dfrac{8}{15}[/tex]
Step by step explanation:
[tex]\text{Given that,}\\\\~~~~~~~\sin \theta = \dfrac{8}{17}\\\\\implies \sin^2 \theta = \dfrac{64}{289}\\\\\implies 1-\cos^2 \theta = \dfrac{64}{289}\\\\\implies \cos^2 \theta = 1-\dfrac{64}{289}\\\\\implies \cos^2 \theta = \dfrac{225}{289}\\\\\implies \cos \theta =\pm\sqrt{\dfrac{225}{289}}\\\\\implies \cos \theta = \pm\dfrac{15}{17}[/tex]
[tex]\text{In quadrant I, all ratios are positive, so,}~ \cos \theta = \dfrac{15}{17}\\\\\text{Now,}\\\\\tan \theta = \dfrac{\sin \theta }{\cos \theta}\\\\\\~~~~~~~~=\dfrac{\tfrac{8}{17}}{\tfrac{15}{17}}\\\\\\~~~~~~~~=\dfrac{8}{17} \times \dfrac{17}{15}\\\\\\~~~~~~~=\dfrac{8}{15}[/tex]
