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 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 )
Sure! I can help you with that.
To find the value of tan(theta) given that csc(theta) = -2 and theta lies in Quadrant III, we can use the reciprocal identities for sine and cosine.
The reciprocal identity for sine is: csc(theta) = 1/sin(theta)
Therefore, we can write: 1/sin(theta) = -2
To solve for sin(theta), we can take the reciprocal of both sides:
sin(theta) = 1/(-2) = -1/2
Since theta lies in Quadrant III, where sine is negative, we can determine the value of cosine using the Pythagorean identity:
sin^2(theta) + cos^2(theta) = 1
((-1/2)^2) + cos^2(theta) = 1
Simplifying the equation:
1/4 + cos^2(theta) = 1
cos^2(theta) = 3/4
Taking the square root of both sides, since cosine is positive in Quadrant III:
cos(theta) = √(3/4) = √3/2
Finally, to find tan(theta), we can use the identity:
tan(theta) = sin(theta) / cos(theta)
Substituting the values we found:
tan(theta) = (-1/2) / (√3/2)
Simplifying:
tan(theta) = -1 / √3 = -√3 / 3
So, the value of tan(theta) when csc(theta) = -2 and theta lies in Quadrant III is -√3 / 3.
I hope this explanation helps! If you have any further questions, feel free to ask.