Find the exact value of each of the remaining trigonometric functions of \( \theta \). \[ \cos \theta=-\frac{7}{25} \text {, } \theta \text { in Quadrant II } \] \( \sin \theta=\square \) (Simplify your answer. Type an exact answer, using radicals as needed. Use integers or the expression.)

To find the sine function, we can use the Pythagorean identity, which states that: \[ \sin^2 \theta + \cos^2 \theta = 1 \] Given that \( \cos \theta = -\frac{7}{25} \), we can substitute that into the identity: \[ \sin^2 \theta + \left(-\frac{7}{25}\right)^2 = 1 \] Calculating \( \left(-\frac{7}{25}\right)^2 \): \[ \sin^2 \theta + \frac{49}{625} = 1 \] Next, we subtract \( \frac{49}{625} \) from both sides: \[ \sin^2 \theta = 1 - \frac{49}{625} = \frac{625 - 49}{625} = \frac{576}{625} \] Now, taking the square root of both sides, we have: \[ \sin \theta = \pm \sqrt{\frac{576}{625}} = \pm \frac{\sqrt{576}}{\sqrt{625}} = \pm \frac{24}{25} \] Since we are in Quadrant II, where sine is positive, we choose the positive value: \[ \sin \theta = \frac{24}{25} \] So, the exact value for \( \sin \theta \) is: \[ \sin \theta = \frac{24}{25} \] Now, let's find the remaining trigonometric functions. 1. **Tangent:** We can use the relationship \( \tan \theta = \frac{\sin \theta}{\cos \theta} \): \[ \tan \theta = \frac{\frac{24}{25}}{-\frac{7}{25}} = -\frac{24}{7} \] 2. **Cosecant:** This is the reciprocal of sine: \[ \csc \theta = \frac{1}{\sin \theta} = \frac{25}{24} \] 3. **Secant:** This is the reciprocal of cosine: \[ \sec \theta = \frac{1}{\cos \theta} = -\frac{25}{7} \] 4. **Cotangent:** This is the reciprocal of tangent: \[ \cot \theta = \frac{1}{\tan \theta} = -\frac{7}{24} \] In summary, the values of the trigonometric functions are: - \( \sin \theta = \frac{24}{25} \) - \( \tan \theta = -\frac{24}{7} \) - \( \csc \theta = \frac{25}{24} \) - \( \sec \theta = -\frac{25}{7} \) - \( \cot \theta = -\frac{7}{24} \)