91Ó°ÊÓ

Given, \(\tan(u) = -2\). Since \(\tan(u) = \frac{\sin(u)}{\cos(u)}\), we have \(-2 = \frac{\sin(u)}{\cos(u)}\). We also know that \(\sin^2(u) + \cos^2(u) = 1\). Now, we can solve for \(\cos(u)\) and \(\sin(u)\) using these equations. 1. First, solve for \(\sin(u)\) or \(\cos(u)\) using the \(\tan(u)\)-equation. For example, we can find \(\sin(u) = -2 \cos(u)\). 2. Substitute the expression for \(\sin(u)\) into the Pythagorean identity \(\sin^2(u) + \cos^2(u) = 1\). 3. Solve the quadratic equation for \(\cos(u)\) to get the result. Since the given interval is in the second quadrant, choose the negative root. 4. Determine \(\sin(u)\) by substituting the value of \(\cos(u)\) back into the expression found in Step 1.

Similar to Step 1, we can find \(\cos(v)\) and \(\sin(v)\) using the information about \(\tan(v)\). Given, \(\tan(v) = -3\). Since \(\tan(v) = \frac{\sin(v)}{\cos(v)}\), we have \(-3 = \frac{\sin(v)}{\cos(v)}\). We also know that \(\sin^2(v) + \cos^2(v) = 1\). Now, we can solve for \(\cos(v)\) and \(\sin(v)\) using these equations. Follow the same procedure as in step 1.

Once we have found the value of \(\cos(v)\) from Step 2, we can determine the value of \(\cos(-v)\) using the property \(\cos(-x) = \cos(x)\). So, \(\cos(-v) = \cos(v)\). Substitute the value of \(\cos(v)\) found in Step 2 to get the final result.