91Ó°ÊÓ

Since we know \(\tan \theta > 0\) and \(\cos \theta < 0\), we can determine the quadrant of \(\theta\). In the second quadrant, sine is positive, cosine is negative, and tangent is positive. In the fourth quadrant, sine is negative, cosine is negative, and tangent is positive. Since \(\tan \theta = 2 > 0\), \(\theta\) is in the second quadrant.

We know \(\tan \theta = \frac{\sin \theta}{\cos \theta} = 2\), and since \(\theta\) is in the second quadrant, \(\cos \theta < 0\). Let's denote the opposite side by \(a\), the adjacent side by \(b\), and the hypotenuse side by \(c\). Then we can write the relation \(\tan \theta = \frac{a}{b} = 2\). Solving for \(a\) gives us \(a = 2b\). Using the Pythagorean identity for right triangles, \(a^2 + b^2 = c^2\), we get \((2b)^2 + b^2 = c^2\), which simplifies to \(5b^2 = c^2\). Now, we can calculate \(\sin \theta = \frac{a}{c} = \frac{2b}{\sqrt{5b^2}} = \frac{2}{\sqrt{5}}\) and \(\cos \theta = \frac{b}{c} = \frac{b}{\sqrt{5b^2}} = -\frac{1}{\sqrt{5}}\) (with the negative sign due to being in the second quadrant).

Now that we have \(\sin \theta\) and \(\cos \theta\), we can find the remaining trigonometric functions: 1. Cosecant (csc): \(\csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{5}}{2}\) 2. Secant (sec): \(\sec \theta = \frac{1}{\cos \theta} = -\sqrt{5}\) 3. Cotangent (cot): \(\cot \theta = \frac{1}{\tan \theta} = \frac{1}{2}\) Thus, the trigonometric functions of \(\theta\) are: 1. \(\sin \theta = \frac{2}{\sqrt{5}}\) 2. \(\cos \theta = -\frac{1}{\sqrt{5}}\) 3. \(\tan \theta = 2\) 4. \(\csc \theta = \frac{\sqrt{5}}{2}\) 5. \(\sec \theta = -\sqrt{5}\) 6. \(\cot \theta = \frac{1}{2}\)