There are several methods to solve this problem.
Geometric
Imagine a right-angle triangle with an angle θ. Since csc(θ)=5, the hypotenuse is 5, its opposite side is 1, and its adjacent side is √52−12=√24=2√6. Thus, tan(θ) is the opposite over the adjacent, or tan(θ)=12√6=√612.
Now, since csc(θ)=5>0, θ is either in quadrant 1 or 2. Because tan(θ) is positive in quadrant 1 and negative in quadrant 2, there are two possible answers: ±√612.
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(θ)=51=5. However, the adjacent side, which is the x-value, becomes negative in quadrant 2 (−√52−12=−√24=−2√6). Thus, tan(θ), being the opposite over the adjacent, has the exact same value but with a negative sign.
Alegbraic
Another method is algebraic:
csc(θ)=5
1sin(θ)=5
sin(θ)=15
θ=arcsin(15)
tan(θ)=tan(arcsin(15))=sin(arcsin(15))cos(arcsin(15))
tan(θ)=15±√1−sin2(arcsin(15))=1±5√1−(15)2
tan(θ)=1±√24=±√612
