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 +
Help please! Thank you!
If csc(theta)= -2 and theta lies in Quadrant III, find tan(theta).
2 answers
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 )
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 )
1 answer