Starting at point \( (1,0) \), you move along a distance \( t \) along a unit circle to terminal point \( P(x, y) \). If \( \sin t=\frac{2}{7} \) and the terminal point is in Quadrant \( I \), find the other 5 trigonometric functions. Express your answer as a reduced fraction, NO DECIMALS! \( \cos t= \) \( \tan t= \) \( \csc t= \) \( \sec t= \)

To find the other trigonometric functions based on the value of \(\sin t = \frac{2}{7}\) while knowing that the terminal point is in Quadrant I, we can proceed as follows: 1. **Finding \(\cos t\):** We know from the Pythagorean identity that: \[ \sin^2 t + \cos^2 t = 1 \] Plugging in the value of \(\sin t\): \[ \left(\frac{2}{7}\right)^2 + \cos^2 t = 1 \] \[ \frac{4}{49} + \cos^2 t = 1 \] \[ \cos^2 t = 1 - \frac{4}{49} = \frac{49}{49} - \frac{4}{49} = \frac{45}{49} \] Taking the square root (since we are in Quadrant I where \(\cos\) is positive): \[ \cos t = \sqrt{\frac{45}{49}} = \frac{\sqrt{45}}{7} = \frac{3\sqrt{5}}{7} \] 2. **Finding \(\tan t\):** Using the relationship \(\tan t = \frac{\sin t}{\cos t}\): \[ \tan t = \frac{\frac{2}{7}}{\frac{3\sqrt{5}}{7}} = \frac{2}{3\sqrt{5}} \] 3. **Finding \(\csc t\):** The cosecant function is the reciprocal of sine: \[ \csc t = \frac{1}{\sin t} = \frac{1}{\frac{2}{7}} = \frac{7}{2} \] 4. **Finding \(\sec t\):** The secant function is the reciprocal of cosine: \[ \sec t = \frac{1}{\cos t} = \frac{1}{\frac{3\sqrt{5}}{7}} = \frac{7}{3\sqrt{5}} \] Now, consolidating all the values: \[ \cos t= \frac{3\sqrt{5}}{7} \] \[ \tan t= \frac{2}{3\sqrt{5}} \] \[ \csc t= \frac{7}{2} \] \[ \sec t= \frac{7}{3\sqrt{5}} \]