Posted by Mandy on Tuesday, February 7, 2012 at 8:55pm.
cscØ = -2
sinØ = -1/2, Ø in III
Ø = 210°
then tan 210° = 1/√3
(I sketched the right-angled 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 )
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