There are several methods to solve this problem.
Geometric
Imagine a right-angle triangle with an angle theta. Since csc(theta)=5, the hypotenuse is 5, its opposite side is 1, and its adjacent side is sqrt(5^2-1^2)=sqrt(24)=2sqrt(6). Thus, tan(theta) is the opposite over the adjacent, or tan(theta)=1/(2sqrt(6))=sqrt(6)/12.
Now, since csc(theta)=5>0, theta is either in quadrant 1 or 2. Because tan(theta) is positive in quadrant 1 and negative in quadrant 2, there are two possible answers: +-sqrt(6)/12.
Another way to realize that there is a second solution is to realize that it is possible to draw the right triangle in the second quadrant. The hypotenuse is 5 (remember that the hypotenuse is always positive). The opposite side, which is the y-value, is still positive (it is still 1). Thus, csc(theta)=5/1=5. However, the adjacent side, which is the x-value, becomes negative in quadrant 2 (-sqrt(5^2-1^2)=-sqrt(24)=-2sqrt(6)). Thus, tan(theta), being the opposite over the adjacent, has the exact same value but with a negative sign.
Alegbraic
Another method is algebraic:
csc(theta)=5
1/sin(theta)=5
sin(theta)=1/5
theta=arcsin(1/5)
tan(theta)=tan(arcsin(1/5))=sin(arcsin(1/5))/cos(arcsin(1/5))
tan(theta)=(1/5)/(+-sqrt(1-sin^2(arcsin(1/5))))=1/(+-5sqrt(1-(1/5)^2))
tan(theta)=1/(+-sqrt(24))=+-sqrt(6)/12
