1. **State the problem:** Given that $\tan \theta = -\frac{5}{12}$ and $\cos \theta$ is negative, find all the trigonometric functions of $\theta$. 2. **Recall the definitions and formulas:** - $\tan \theta = \frac{\sin \theta}{\cos \theta}$ - $\sin^2 \theta + \cos^2 \theta = 1$ 3. **Determine the quadrant:** Since $\tan \theta$ is negative and $\cos \theta$ is negative, $\theta$ lies in the third quadrant (where sine and cosine are negative, tangent is positive) or second quadrant (where sine is positive, cosine negative, tangent negative). But tangent is negative, so $\theta$ is in the second quadrant. 4. **Find $\sin \theta$ and $\cos \theta$ using $\tan \theta$:** Let $\tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{5}{12}$. Assume $\sin \theta = -5k$ and $\cos \theta = 12k$ for some $k$ (signs to be adjusted based on quadrant). Since $\cos \theta$ is negative, $\cos \theta = -12k$. Since $\tan \theta$ is negative, and $\tan \theta = \frac{\sin \theta}{\cos \theta} = -\frac{5}{12}$, and $\cos \theta$ is negative, $\sin \theta$ must be positive. So $\sin \theta = 5k$, $\cos \theta = -12k$. 5. **Use Pythagorean identity:** $$\sin^2 \theta + \cos^2 \theta = 1$$ $$ (5k)^2 + (-12k)^2 = 1 $$ $$ 25k^2 + 144k^2 = 1 $$ $$ 169k^2 = 1 $$ $$ k^2 = \frac{1}{169} $$ $$ k = \frac{1}{13} $$ 6. **Calculate $\sin \theta$ and $\cos \theta$:** $$ \sin \theta = 5k = 5 \times \frac{1}{13} = \frac{5}{13} $$ $$ \cos \theta = -12k = -12 \times \frac{1}{13} = -\frac{12}{13} $$ 7. **Calculate $\sec \theta$, $\csc \theta$, and $\cot \theta$:** $$ \sec \theta = \frac{1}{\cos \theta} = \frac{1}{-\frac{12}{13}} = -\frac{13}{12} $$ $$ \csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{5}{13}} = \frac{13}{5} $$ $$ \cot \theta = \frac{1}{\tan \theta} = \frac{1}{-\frac{5}{12}} = -\frac{12}{5} $$ **Final answers:** $$ \sin \theta = \frac{5}{13}, \quad \cos \theta = -\frac{12}{13}, \quad \tan \theta = -\frac{5}{12} $$ $$ \csc \theta = \frac{13}{5}, \quad \sec \theta = -\frac{13}{12}, \quad \cot \theta = -\frac{12}{5} $$