1. The problem asks for the reciprocal of $\sin\theta$. The reciprocal of a trigonometric function $f(\theta)$ is $\frac{1}{f(\theta)}$. For $\sin\theta$, the reciprocal is $\csc\theta$. So, the answer is $\csc\theta$. 2. Next, we determine the quadrant of angle $\theta$ given $\cos\theta < 0$ and $\sin\theta > 0$. Recall the signs of sine and cosine in each quadrant: - Quadrant I: $\sin > 0$, $\cos > 0$ - Quadrant II: $\sin > 0$, $\cos < 0$ - Quadrant III: $\sin < 0$, $\cos < 0$ - Quadrant IV: $\sin < 0$, $\cos > 0$ Since $\sin\theta > 0$ and $\cos\theta < 0$, $\theta$ lies in Quadrant II. 3. Given $\tan\theta = -\frac{2}{3}$ and $\sec\theta > 0$, find $\csc\theta$. Step 1: Recall that $\tan\theta = \frac{\sin\theta}{\cos\theta} = -\frac{2}{3}$. Step 2: Since $\sec\theta = \frac{1}{\cos\theta} > 0$, $\cos\theta > 0$. Tangent is negative and cosine positive, so $\theta$ is in Quadrant IV. Step 3: Use the Pythagorean identity: $$1 + \tan^2\theta = \sec^2\theta$$ Calculate $\sec\theta$: $$1 + \left(-\frac{2}{3}\right)^2 = \sec^2\theta$$ $$1 + \frac{4}{9} = \sec^2\theta$$ $$\frac{13}{9} = \sec^2\theta$$ $$\sec\theta = \frac{\sqrt{13}}{3}$$ (positive as given) Step 4: Find $\cos\theta$: $$\cos\theta = \frac{1}{\sec\theta} = \frac{3}{\sqrt{13}}$$ Step 5: Find $\sin\theta$ using $\tan\theta = \frac{\sin\theta}{\cos\theta}$: $$-\frac{2}{3} = \frac{\sin\theta}{\frac{3}{\sqrt{13}}}$$ $$\sin\theta = -\frac{2}{3} \times \frac{3}{\sqrt{13}} = -\frac{2}{\sqrt{13}}$$ Step 6: Calculate $\csc\theta = \frac{1}{\sin\theta}$: $$\csc\theta = \frac{1}{-\frac{2}{\sqrt{13}}} = -\frac{\sqrt{13}}{2}$$ Therefore, the value of $\csc\theta$ is $-\frac{\sqrt{13}}{2}$.