\( \begin{array}{llll}\text { 1) } \frac{4}{3} & \text { 2) }-\frac{\sqrt{7}}{4} & \text { 3) } \frac{\sqrt{7}}{3} & \text { 4) }-\frac{\sqrt{7}}{3} \text { If } \sin \theta=-\frac{3}{5} \text { and } \cos \theta>0 \text {, what is the value of } \tan \theta \text { ? } \begin{array}{llll}\text { 1) } \frac{3}{4} & \text { 2) }-\frac{3}{4} & \text { 3) } \frac{4}{3} & \text { 4) }-\frac{4}{3}\end{array}\end{array} \)

To find \( \tan \theta \), we use the relationship \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). Given that \( \sin \theta = -\frac{3}{5} \) and \( \cos \theta > 0 \), we need to find \( \cos \theta \). Applying the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \), we get: \[ \left(-\frac{3}{5}\right)^2 + \cos^2 \theta = 1 \Rightarrow \frac{9}{25} + \cos^2 \theta = 1 \Rightarrow \cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25} \] Since \( \cos \theta > 0 \), we take the positive root: \[ \cos \theta = \frac{4}{5} \] Now, we can calculate \( \tan \theta \): \[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{-\frac{3}{5}}{\frac{4}{5}} = -\frac{3}{4} \] So the value of \( \tan \theta \) is \( -\frac{3}{4} \), which corresponds to option 2. Understanding the relationship between sine, cosine, and tangent can be additionally exciting! Did you know that in the unit circle, the sine function corresponds to the y-coordinate while cosine corresponds to the x-coordinate? This makes visualizing these trigonometric functions quite dynamic, especially in determining in which quadrant the angle lies. For practical applications, the tangent function often emerges in real-life scenarios such as calculating slopes or heights from angles, such as in architecture or engineering. A right triangle can help visualize real-world problems: for instance, if you're trying to find the height of a tree by measuring the angle of elevation from a certain distance away, you're applying the tangent function directly!