Given that \(\cos(\theta) = -\frac{3}{5}\) and \(\theta\) is in the second quadrant, find the values of \(\sin(\theta)\), \(\tan(\theta)\), \(\csc(\theta)\), \(\sec(\theta)\), and \(\cot(\theta)\).

Step 1 ：Step 1: We know that \(\cos(\theta)\) is negative in the second quadrant, and the value of \(\sin(\theta)\) is positive. We can use the Pythagorean identity \(\sin^2(\theta) + \cos^2(\theta) = 1\) to find \(\sin(\theta)\). So \(\sin^2(\theta) = 1 - \cos^2(\theta) = 1 - \left(-\frac{3}{5}\right)^2 = \frac{16}{25}\), thus \(\sin(\theta) = \sqrt{\frac{16}{25}} = \frac{4}{5}\)

Step 2 ：Step 2: We can find \(\tan(\theta)\) using the identity \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\). So \(\tan(\theta) = \frac{\frac{4}{5}}{-\frac{3}{5}} = -\frac{4}{3}\)

Step 3 ：Step 3: The reciprocal of \(\sin(\theta)\) is \(\csc(\theta)\), so \(\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{\frac{4}{5}} = \frac{5}{4}\)

Step 4 ：Step 4: The reciprocal of \(\cos(\theta)\) is \(\sec(\theta)\), so \(\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{3}{5}} = -\frac{5}{3}\)

Step 5 ：Step 5: The reciprocal of \(\tan(\theta)\) is \(\cot(\theta)\), so \(\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{-\frac{4}{3}} = -\frac{3}{4}\)