Use the given information to find the exact value of a. $\sin 2 \theta, b \cdot \cos 2 \theta$, and $c . \tan 2 \theta$. $\cos \theta=\frac{6}{10}, \theta$ lies in quadrant IV

Step 1 ：Given that \(\cos \theta = \frac{6}{10}\) and \(\theta\) lies in quadrant IV, we can find the value of \(\sin \theta\) using the Pythagorean identity \(\sin^2 \theta + \cos^2 \theta = 1\). Since \(\theta\) is in the fourth quadrant, \(\sin \theta\) will be negative, so \(\sin \theta = -\sqrt{1 - \cos^2 \theta} = -0.8\).

Step 2 ：We can then use the double angle formulas to find the values of \(\sin 2 \theta\), \(\cos 2 \theta\), and \(\tan 2 \theta\). The double angle formulas are \(\sin 2 \theta = 2 \sin \theta \cos \theta\), \(\cos 2 \theta = \cos^2 \theta - \sin^2 \theta\), and \(\tan 2 \theta = \frac{\sin 2 \theta}{\cos 2 \theta}\).

Step 3 ：Using these formulas, we find that \(\sin 2 \theta = 2 \cdot -0.8 \cdot 0.6 = -0.96\), \(\cos 2 \theta = 0.6^2 - (-0.8)^2 = -0.28\), and \(\tan 2 \theta = \frac{-0.96}{-0.28} = 3.43\).

Step 4 ：So, the exact values of \(\sin 2 \theta\), \(b \cdot \cos 2 \theta\), and \(c \cdot \tan 2 \theta\) are \(\boxed{-0.96}\), \(\boxed{-0.28b}\), and \(\boxed{3.43c}\) respectively.