If \( \cos 34^{\circ}=t \), express each of the following in terms of \( t \) : \( \begin{array}{ll}\text { 3.1.1 } & \sin 34^{\circ} \\ \text { 3.1.2 } & \tan 146^{\circ}\end{array} \)

We are given \[ \cos 34^\circ = t. \] **Step 1. Express \(\sin 34^\circ\) in terms of \(t\):** We use the Pythagorean identity \[ \sin^2 34^\circ + \cos^2 34^\circ = 1. \] Substitute \(\cos 34^\circ = t\): \[ \sin^2 34^\circ + t^2 = 1. \] Thus, \[ \sin^2 34^\circ = 1 - t^2. \] Since \(34^\circ\) is in the first quadrant, \(\sin 34^\circ\) is positive. Therefore, \[ \sin 34^\circ = \sqrt{1-t^2}. \] **Step 2. Express \(\tan 146^\circ\) in terms of \(t\):** Recognize that \[ 146^\circ = 180^\circ - 34^\circ. \] The tangent identity for supplementary angles is \[ \tan(180^\circ - \theta) = -\tan \theta. \] Thus, \[ \tan 146^\circ = -\tan 34^\circ. \] Now, by definition \[ \tan 34^\circ = \frac{\sin 34^\circ}{\cos 34^\circ}. \] Substitute the expressions we have: \[ \tan 34^\circ = \frac{\sqrt{1-t^2}}{t}. \] Therefore, \[ \tan 146^\circ = -\frac{\sqrt{1-t^2}}{t}. \] **Final Answers:** - \(\sin 34^\circ = \sqrt{1-t^2}\) - \(\tan 146^\circ = -\frac{\sqrt{1-t^2}}{t}\)