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#