Find the remaining trigonometric ratios of θ if sin(θ) = -1/6 and θ is not in QIII
cos
tan
cot
csc
sec
so theta is in quadrant 4
triangle (-1)^2 + x^2 = 6^2
x = sqrt 35
cos = (1/6)sqrt 35
tan = -1/sqrt 35
etc
To find the remaining trigonometric ratios of θ, we need to first determine the quadrant where θ lies, since the given information states that θ is not in QIII.
If sin(θ) = -1/6, we know that sin(θ) is negative in QIII and QIV. However, since θ is not in QIII, it must be in QIV.
In QIV, we have a negative value for sin(θ), which means that cos(θ) is positive.
Now, let's use the Pythagorean identity to find the value of cos(θ):
sin^2(θ) + cos^2(θ) = 1
(-1/6)^2 + cos^2(θ) = 1
1/36 + cos^2(θ) = 1
cos^2(θ) = 1 - 1/36
cos^2(θ) = 35/36
Since cos(θ) is positive in QIV, we take the positive square root of 35/36:
cos(θ) = √(35/36) = √35/6
Now that we have the value of cos(θ), we can find the remaining trigonometric ratios.
1. tan(θ): tan(θ) = sin(θ)/cos(θ)
tan(θ) = (-1/6) / (√35/6)
tan(θ) = -1/√35
2. cot(θ): cot(θ) = 1/tan(θ)
cot(θ) = 1 / (-1/√35)
cot(θ) = -√35
3. csc(θ): csc(θ) = 1/sin(θ)
csc(θ) = 1 / (-1/6)
csc(θ) = -6
4. sec(θ): sec(θ) = 1/cos(θ)
sec(θ) = 1 / (√35/6)
sec(θ) = 6/√35
Hence, the remaining trigonometric ratios for θ with sin(θ) = -1/6 and θ not in QIII are:
cos(θ) = √35/6
tan(θ) = -1/√35
cot(θ) = -√35
csc(θ) = -6
sec(θ) = 6/√35