Answer:
[tex]\large\boxed{\sec\theta=\dfrac{7}{\sqrt{24}},\ and\ \tan\theta=-\dfrac{5}{\sqrt{24}}}\\\boxed{\cos\theta=\dfrac{\sqrt{24}}{7},\ and\ \tan\theta=\dfrac{5}{\sqrt{24}}}[/tex]
Step-by-step explanation:
[tex]\text{Use:}\\\\\sin^2\theta+\cos^2\theta=1\to\cos\theta=\pm\sqrt{1-\sin^2\theta}\\\\\tan\theta=\dfrac{\sin\theta}{\cos\theta}\\\\\sec\theta=\dfrac{1}{\cos\theta}\\\\==========================\\\\\sin\theta=-\dfrac{5}{7}\\\\\cos\theta=\pm\sqrt{1-\left(-\frac{5}{7}\right)^2}=\pm\sqrt{1-\frac{25}{49}}=\pm\sqrt{\frac{24}{49}}=\pm\dfrac{\sqrt{24}}{\sqrt{49}}=\pm\dfrac{\sqrt{24}}{7}\\\\\sec\theta=\pm\dfrac{1}{\frac{\sqrt{24}}{7}}=\pm\dfrac{7}{\sqrt{24}}[/tex]
[tex]\tan\theta=\pm\dfrac{-\frac{5}{7}}{\frac{\sqrt{24}}{7}}=\mp\dfrac{5}{7\!\!\!\!\diagup_1}\cdot\dfrac{7\!\!\!\!\diagup^1}{\sqrt{24}}=\mp\dfrac{5}{\sqrt{24}}[/tex]
[tex]\sin\theta=-\dfrac{5}{7}<0\to\text{III or IV Quadrant}\\\\\text{Look at the picture}\\\\\text{Quadrant III}:\\\\\sin<0,\ \cos<0,\ \sec<0,\ \tan>0\\\\\text{Quadrant IV}:\\\\\sin<0,\ \cos>0,\ \sec>0,\ \tan<0[/tex]
