Posted by **Mandy** on Tuesday, February 7, 2012 at 8:55pm.

Help please! Thank you!

If csc(theta)= -2 and theta lies in Quadrant III, find tan(theta).

- Trig -
**Reiny**, Tuesday, February 7, 2012 at 11:06pm
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 +

- Trig -
**Bosnian**, Wednesday, February 8, 2012 at 12:41am
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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