91Ó°ÊÓ

Since \(\sin t = -2/3\) and \(\sec t > 0\), we can find in which quadrant the angle t lies. Since the sine is negative and the secant (reciprocal of cosine) is positive, the angle t must be in the fourth quadrant, where cosine is positive and sine is negative. Now we can use the Pythagorean identity: \(\sin^2 t + \cos^2 t = 1\), to find the value of cosine: $$\cos^2 t = 1 - \sin^2 t = 1 - (-\frac{2}{3})^2 = 1 - \frac{4}{9} = \frac{5}{9}$$ Since t is in the fourth quadrant and cosine is positive there, we have: $$\cos t = \sqrt{\frac{5}{9}} = \frac{\sqrt{5}}{3}$$

Now that we have the values of sine and cosine, we can find the values of the other four trigonometric functions by using their relationships with sine and cosine: 1. Tangent: \(\tan t = \frac{\sin t}{\cos t} = \frac{-2/3}{\sqrt{5}/3} = -\frac{2}{\sqrt{5}} = -\frac{2\sqrt{5}}{5}\) (by rationalizing the denominator) 2. Cotangent: \(\cot t = \frac{1}{\tan t} = -\frac{\sqrt{5}}{2}\) 3. Cosecant: \(\csc t = \frac{1}{\sin t} = -\frac{3}{2}\) 4. Secant: \(\sec t = \frac{1}{\cos t} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}\) (by rationalizing the denominator) So we have found the values of all six trigonometric functions at t: 1. \(\sin t = -\frac{2}{3}\) 2. \(\cos t = \frac{\sqrt{5}}{3}\) 3. \(\tan t = -\frac{2\sqrt{5}}{5}\) 4. \(\cot t = -\frac{\sqrt{5}}{2}\) 5. \(\csc t = -\frac{3}{2}\) 6. \(\sec t = \frac{3\sqrt{5}}{5}\)