Find the remaining five trigonometric functions of \( \theta \) satisfying the conditions. (Himts Draw a triangle in the appropriate quadrant. Pay attention to signs! \[ 15, \tan \theta=\frac{7}{3}, \cos \theta<0 \]

Let the terminal side of \( \theta \) lie in Quadrant III so that both \( \sin\theta \) and \( \cos\theta \) are negative and \( \tan\theta \) is positive. We are given \[ \tan\theta = \frac{7}{3} \quad \text{and} \quad r=15 \] with the extra information that \( \cos\theta<0 \). 1. **Assign side lengths** Since \[ \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{3}, \] we let \[ \text{opposite} = 7t \quad \text{and} \quad \text{adjacent} = 3t, \] where \( t>0 \). Because \( \theta \) is in the third quadrant, we have \[ x = -3t \quad \text{and} \quad y = -7t. \] 2. **Relate to the hypotenuse** Using the Pythagorean theorem, \[ r = \sqrt{(-3t)^2 + (-7t)^2} = \sqrt{9t^2+49t^2} = \sqrt{58t^2} = t\sqrt{58}. \] Given \( r = 15 \), we have \[ t\sqrt{58} = 15 \quad \Longrightarrow \quad t = \frac{15}{\sqrt{58}}. \] 3. **Find the sine and cosine** The definitions of sine and cosine give: \[ \cos\theta = \frac{x}{r} = \frac{-3t}{15} = -\frac{3}{15}\, t\times\frac{15}{t} \quad \text{(substituting }t=\frac{15}{\sqrt{58}}\text{)}, \] so \[ \cos\theta = -\frac{3t}{15} = -\frac{3\left(\frac{15}{\sqrt{58}}\right)}{15} = -\frac{3}{\sqrt{58}}. \] Similarly, \[ \sin\theta = \frac{y}{r} = \frac{-7t}{15} = -\frac{7\left(\frac{15}{\sqrt{58}}\right)}{15} = -\frac{7}{\sqrt{58}}. \] 4. **Confirm the tangent** We check: \[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{-\frac{7}{\sqrt{58}}}{-\frac{3}{\sqrt{58}}} = \frac{7}{3}, \] which agrees with the given information. 5. **Find the reciprocal functions** The remaining functions are the reciprocals: \[ \sec\theta = \frac{1}{\cos\theta} = -\frac{\sqrt{58}}{3}, \] \[ \csc\theta = \frac{1}{\sin\theta} = -\frac{\sqrt{58}}{7}, \] \[ \cot\theta = \frac{1}{\tan\theta} = \frac{3}{7}. \] 6. **Final Answers** \[ \sin\theta = -\frac{7}{\sqrt{58}}, \quad \cos\theta = -\frac{3}{\sqrt{58}}, \quad \tan\theta = \frac{7}{3}, \] \[ \csc\theta = -\frac{\sqrt{58}}{7}, \quad \sec\theta = -\frac{\sqrt{58}}{3}, \quad \cot\theta = \frac{3}{7}. \] These are the remaining five trigonometric functions satisfying the given conditions.