Find the exact value of each of the remaining trigonometric functions of \( \theta \). \[ \tan \theta=-\frac{2}{5}, \csc \theta>0 \]

We are given \[ \tan\theta = -\frac{2}{5} \quad \text{and} \quad \csc\theta > 0. \] **Step 1. Determine the Quadrant** Since \[ \csc\theta = \frac{1}{\sin\theta} > 0, \] we have \(\sin\theta > 0\). Also, \[ \tan\theta = \frac{\sin\theta}{\cos\theta} = -\frac{2}{5} \] is negative. Given that \(\sin\theta > 0\), it follows that \(\cos\theta\) must be negative. Hence, \(\theta\) is in the second quadrant. **Step 2. Set up a Right Triangle** We express the tangent ratio as \[ \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{2}{5}. \] Since \(\tan\theta\) is negative in the second quadrant, we take the numerator (opposite side) positive and the denominator (adjacent side) negative. Thus, let: - Opposite side \(= 2\), - Adjacent side \(= -5\). **Step 3. Find the Hypotenuse** Using the Pythagorean theorem, the hypotenuse \(r\) is: \[ r = \sqrt{(\text{opposite})^2 + (\text{adjacent})^2} = \sqrt{2^2 + (-5)^2} = \sqrt{4 + 25} = \sqrt{29}. \] **Step 4. Compute the Trigonometric Functions** 1. **Sine:** \[ \sin\theta = \frac{\text{opposite}}{r} = \frac{2}{\sqrt{29}} = \frac{2\sqrt{29}}{29} \quad (\text{after rationalizing}). \] 2. **Cosine:** \[ \cos\theta = \frac{\text{adjacent}}{r} = \frac{-5}{\sqrt{29}} = -\frac{5\sqrt{29}}{29}. \] 3. **Tangent:** (given) \[ \tan\theta = -\frac{2}{5}. \] 4. **Cosecant:** \[ \csc\theta = \frac{1}{\sin\theta} = \frac{\sqrt{29}}{2}. \] 5. **Secant:** \[ \sec\theta = \frac{1}{\cos\theta} = -\frac{\sqrt{29}}{5}. \] 6. **Cotangent:** \[ \cot\theta = \frac{1}{\tan\theta} = -\frac{5}{2}. \] **Final Answers:** \[ \sin\theta = \frac{2\sqrt{29}}{29}, \quad \cos\theta = -\frac{5\sqrt{29}}{29}, \quad \tan\theta = -\frac{2}{5}, \] \[ \csc\theta = \frac{\sqrt{29}}{2}, \quad \sec\theta = -\frac{\sqrt{29}}{5}, \quad \cot\theta = -\frac{5}{2}. \]