find the exact value of \\( \\csc \\theta \\), given that \\( \\cot \\theta=-\\frac{1}{4} \\) and \\(…

Explanation:

Step1: Use the Pythagorean identity

We know the identity (1+\cot^{2}\theta=\csc^{2}\theta). Given (\cot\theta =-\frac{1}{4}), substitute it into the identity: (1+\left(-\frac{1}{4}\right)^{2}=\csc^{2}\theta).

Step2: Simplify the left - hand side

Calculate (\left(-\frac{1}{4}\right)^{2}=\frac{1}{16}), then (1+\frac{1}{16}=\csc^{2}\theta). (1+\frac{1}{16}=\frac{16 + 1}{16}=\frac{17}{16}), so (\csc^{2}\theta=\frac{17}{16}).

Step3: Solve for (\csc\theta)

Take the square root of both sides: (\csc\theta=\pm\sqrt{\frac{17}{16}}=\pm\frac{\sqrt{17}}{4}). Since (\theta) is in quadrant IV, and (\csc\theta=\frac{1}{\sin\theta}), and (\sin\theta<0) in quadrant IV, so (\csc\theta<0).

Answer:

(\csc\theta =-\frac{\sqrt{17}}{4})