Consider angle \( \alpha \), where \( 0<\alpha<180^{\circ} \) and \( \tan \theta=-\frac{8}{3} \). Find the value of all six trigonometric functions for the angle \( \theta \). Simplify radicals and rationalize denominatorst if necessary.

To find the six trigonometric functions for the angle \( \theta \) given that \( \tan \theta = -\frac{8}{3} \), we start by identifying the quadrant in which the angle lies. Since the tangent is negative, \( \theta \) must be in either the second quadrant (where sine is positive and cosine is negative) or the fourth quadrant (where sine is negative and cosine is positive). We will assume \( \theta \) is in the second quadrant for this solution. First, we can represent the tangent function in terms of the opposite and adjacent sides of a right triangle: \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = -\frac{8}{3}. \] We choose the opposite side to be \( -8 \) and the adjacent side to be \( 3 \) (the negative sign indicates the opposite side is in the negative direction for the fourth quadrant). To find the hypotenuse, we use the Pythagorean theorem: \[ r = \sqrt{(-8)^2 + 3^2} = \sqrt{64 + 9} = \sqrt{73}. \] Now we can find the values of the six trigonometric functions: 1. **Sine**: \[ \sin \theta = \frac{\text{opposite}}{r} = \frac{-8}{\sqrt{73}}. \] 2. **Cosine**: \[ \cos \theta = \frac{\text{adjacent}}{r} = \frac{3}{\sqrt{73}}. \] 3. **Tangent**: \[ \tan \theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{-8}{3}, \] which is already given. 4. **Cosecant** (the reciprocal of sine): \[ \csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{73}}{-8} = -\frac{\sqrt{73}}{8}. \] 5. **Secant** (the reciprocal of cosine): \[ \sec \theta = \frac{1}{\cos \theta} = \frac{\sqrt{73}}{3}. \] 6. **Cotangent** (the reciprocal of tangent): \[ \cot \theta = \frac{1}{\tan \theta} = -\frac{3}{8}. \] Combining all of these, we find: \[ \sin \theta = -\frac{8}{\sqrt{73}}, \quad \cos \theta = \frac{3}{\sqrt{73}}, \quad \tan \theta = -\frac{8}{3}, \] \[ \csc \theta = -\frac{\sqrt{73}}{8}, \quad \sec \theta = \frac{\sqrt{73}}{3}, \quad \cot \theta = -\frac{3}{8}. \]