Trig
posted by Mandy .
Help please! Thank you!
If csc(theta)= 2 and theta lies in Quadrant III, find tan(theta).

cscØ = 2
sinØ = 1/2, Ø in III
Ø = 210°
then tan 210° = 1/√3
(I sketched the rightangled triangle in quad III with sides
1  √3  2, for which you should know the ratios.
from the CAST rule I knew that the tangent in III is + 
In Quadrant III
sin ( theta ) , cos ( theta ) , sec ( theta ) and csc( theta ) are negative.
tan ( theta ) = sin ( theta ) / cos ( theta )
are positive.
tan ( theta ) = + OR 1 / sqrt [ csc ( theta ) ^ 2  1 ]
tan ( theta ) = + OR 1 / sqrt [ ( 2 ) ^ 2  1 ]
tan ( theta ) = + OR 1 / sqrt ( 4  1 )
tan ( theta ) = + OR 1 / sqrt ( 3 )
In Quadrant III
tan ( theta ) are positive so:
tan ( theta ) = 1 / sqrt ( 3 )