If cos ⁡θ=−8/17, and 180°<θ<270°, what is tan ⁡θ?

Answer:[tex]\large\boxed{\tan\theta=\dfrac{8}{15}}[/tex]Step-by-step explanation:[tex]180^o<\theta<270^o\to\bold{Quadrant\ III}-\text{look at the picture}\\\\\tan\theta>0.\\\\\text{We have}\ \cos\theta=-\dfrac{8}{17}.\\\\\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\\text{Use}\ \sin^2\theta+\cos^2\theta=1:\\\\\sin^2\theta+\left(-\dfrac{8}{17}\right)^2=1\\\\\sin^2\theta+\dfrac{64}{289}=1\qquad\text{subtract}\ \dfrac{64}{289}\ \text{from both sides}\\\\\sin^2\theta=\dfrac{289-64}{289}\\\\\sin^2\theta=\dfrac{225}{289}\to\sin\theta=\pm\sqrt{\dfrac{225}{289}}[/tex][tex]\sin\theta=\pm\dfrac{\sqrt{225}}{\sqrt{289}}\to\sin\theta=\pm\dfrac{15}{17}\\\\\bold{Quadrant\ III}\to\sin\theta<0\to\sin\theta=-\dfrac{15}{17}\\\\\text{Substitute to the formula of a tangent}:\\\\\tan\theta=\dfrac{-\frac{8}{17}}{-\frac{15}{17}}=\dfrac{8}{17\!\!\!\!\!\diagup}\cdot\dfrac{17\!\!\!\!\!\diagup}{15}=\dfrac{8}{15}[/tex]