Given that $tan(\alpha) = -\frac{4}{3}$ and $\alpha$ is in the second quadrant, find the value of $cos(\alpha)$.

1. Problem Description

Given that $tan(\alpha) = -\frac{4}{3}$ and $\alpha$ is in the second quadrant, find the value of $cos(\alpha)$.

2. Solution Steps

We know that $tan(\alpha) = \frac{sin(\alpha)}{cos(\alpha)}$.

Also, we know that $sin^2(\alpha) + cos^2(\alpha) = 1$.

Since $tan(\alpha) = -\frac{4}{3}$, we can write $sin(\alpha) = -\frac{4}{3} cos(\alpha)$.

Substituting this into the Pythagorean identity:

$(-\frac{4}{3} cos(\alpha))^2 + cos^2(\alpha) = 1$

$\frac{16}{9} cos^2(\alpha) + cos^2(\alpha) = 1$

$(\frac{16}{9} + 1) cos^2(\alpha) = 1$

$(\frac{16+9}{9}) cos^2(\alpha) = 1$

$\frac{25}{9} cos^2(\alpha) = 1$

$cos^2(\alpha) = \frac{9}{25}$

$cos(\alpha) = \pm \sqrt{\frac{9}{25}} = \pm \frac{3}{5}$

Since $\alpha$ is in the second quadrant, $cos(\alpha)$ is negative. Therefore, $cos(\alpha) = -\frac{3}{5}$.

3. Final Answer

The value of $cos(\alpha)$ is $-\frac{3}{5}$.

The correct option is (B).