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

To find the value of \( \sin x \) given that \( \cos x = -\frac{4}{5} \) and \( \tan x > 0 \), we can first recall the identity \( \sin^2 x + \cos^2 x = 1 \). Starting with \( \cos x = -\frac{4}{5} \): \[ \sin^2 x + \left(-\frac{4}{5}\right)^2 = 1 \] Calculating \( \left(-\frac{4}{5}\right)^2 \): \[ \sin^2 x + \frac{16}{25} = 1 \] Now, subtract \( \frac{16}{25} \) from both sides: \[ \sin^2 x = 1 - \frac{16}{25} = \frac{25}{25} - \frac{16}{25} = \frac{9}{25} \] Taking the square root of both sides: \[ \sin x = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5} \] Since the problem states that \( \tan x > 0 \), we consider the signs of \( \sin x \) and \( \cos x \). Given \( \cos x < 0 \) and \( \tan x = \frac{\sin x}{\cos x} > 0 \), we must have \( \sin x < 0 \). Thus: \[ \sin x = -\frac{3}{5} \] Therefore, the value of \( \sin x \) is \( -\frac{3}{5} \). The correct answer is: **4) \(-\frac{3}{5}\)**