1. **Problem Statement:** Given $\tan \theta = -\frac{12}{5}$ and $\sin \theta > 0$, find the quadrant of $\theta$, then find $\sin \theta$, $\cos \theta$, $\sec \theta$, $\csc \theta$, and $\cot \theta$ in decimal form. 2. **Determine the quadrant:** - $\tan \theta = \frac{\sin \theta}{\cos \theta}$. - Since $\tan \theta$ is negative and $\sin \theta > 0$, $\cos \theta$ must be negative. - $\sin \theta > 0$ means $\theta$ is in either Quadrant I or II. - $\cos \theta < 0$ means $\theta$ is in Quadrant II or III. - The only quadrant satisfying both is **Quadrant II**. 3. **Find $\sin \theta$ and $\cos \theta$:** - Use the identity $\tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{12}{5}$. - Let $\sin \theta = y$ and $\cos \theta = x$. - Then $\frac{y}{x} = -\frac{12}{5}$. - Since $\theta$ is in Quadrant II, $y > 0$ and $x < 0$. - Use Pythagorean identity: $\sin^2 \theta + \cos^2 \theta = 1$. 4. **Express $y$ in terms of $x$:** $$ y = -\frac{12}{5} x $$ 5. **Substitute into Pythagorean identity:** $$ y^2 + x^2 = 1 $$ $$ \left(-\frac{12}{5} x\right)^2 + x^2 = 1 $$ $$ \frac{144}{25} x^2 + x^2 = 1 $$ $$ \left(\frac{144}{25} + 1\right) x^2 = 1 $$ $$ \frac{169}{25} x^2 = 1 $$ $$ x^2 = \frac{25}{169} $$ 6. **Find $x$ (cosine):** - Since $x < 0$ in Quadrant II, $$ x = -\frac{5}{13} \approx -0.3846 $$ 7. **Find $y$ (sine):** $$ y = -\frac{12}{5} x = -\frac{12}{5} \times \left(-\frac{5}{13}\right) = \frac{12}{13} \approx 0.9231 $$ 8. **Find $\sec \theta$:** $$ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{5}{13}} = -\frac{13}{5} = -2.6 $$ 9. **Find $\csc \theta$:** $$ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{12}{13}} = \frac{13}{12} \approx 1.0833 $$ 10. **Find $\cot \theta$:** $$ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-\frac{12}{5}} = -\frac{5}{12} \approx -0.4167 $$ **Final answers:** - Quadrant: II - $\sin \theta \approx 0.9231$ - $\cos \theta \approx -0.3846$ - $\sec \theta \approx -2.6$ - $\csc \theta \approx 1.0833$ - $\cot \theta \approx -0.4167$