If \( \theta \) is an angle with terminal side in the \( 3^{\text {rd }} \) quadrant such that \( \cot \theta=1 \), then \( \sin \theta \) is euqal to : a) \( \sqrt{2} \) b) \( -\frac{1}{\sqrt{2}} \) c) 1 d) \( \frac{1}{\sqrt{2}} \)

To find \( \sin \theta \) given that \( \cot \theta = 1 \) and that \( \theta \) is in the third quadrant, we start by recalling the definition of cotangent: \[ \cot \theta = \frac{\cos \theta}{\sin \theta} \] Since \( \cot \theta = 1 \), we have: \[ \frac{\cos \theta}{\sin \theta} = 1 \] This implies: \[ \cos \theta = \sin \theta \] In the third quadrant, both sine and cosine are negative. Therefore, we can express \( \sin \theta \) and \( \cos \theta \) as: \[ \sin \theta = -x \quad \text{and} \quad \cos \theta = -x \] for some positive \( x \). Since \( \sin \theta = \cos \theta \), we can set: \[ \sin \theta = \cos \theta = -x \] Using the Pythagorean identity: \[ \sin^2 \theta + \cos^2 \theta = 1 \] we substitute \( \sin \theta \) and \( \cos \theta \): \[ (-x)^2 + (-x)^2 = 1 \] This simplifies to: \[ 2x^2 = 1 \] Solving for \( x^2 \): \[ x^2 = \frac{1}{2} \] Taking the square root gives: \[ x = \frac{1}{\sqrt{2}} \] Thus, we have: \[ \sin \theta = -x = -\frac{1}{\sqrt{2}} \] Therefore, the value of \( \sin \theta \) is: \[ \sin \theta = -\frac{1}{\sqrt{2}} \] The correct answer is: \[ \boxed{-\frac{1}{\sqrt{2}}} \]