Find the values of the six trigonometric functions of θ with the given constraint. Function Value sec θ = -2 Constraint sin θ < 0

Solution:

Now since sec θ = -2, it implies 

1/cos θ  = -2

Or 

Cos  θ  = - 1/2

Sin θ < 0 implies that either it is the 3rd or the 4th quadrant we are dealing with. Since Cos θ is negative we can conclude that we are operating in the third quadrant.

Since Cos  θ  = - 1/2 ,

Base = OB = 1

Hypotenuse = OA =2 

Using the Pythagorean relationship for the right angled triangle we get

OA² = OB² +  AB²

2²  =  1² +  AB²

AB² =  4 - 1 = 3

AB = √3

Therefore the six basic trigonometric ratios are::

Sin θ = Perpendicular(AB)/Hypotenuse(OA) = -√3/2

Cos θ = Base (OB)/Hypotenuse(OA) = -1/2 

Tan θ = Perpendicular(AB)/Base(OB) = -√3/-1 = √3

Cosec θ = Hypotenuse/Perpendicular = -2/√3

Sec θ = Hypotenuse/Base = 2/-1 = -2

Cot θ = Base/ Perpendicular  = -1/-√3 = 1/-√3

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Find the values of the six trigonometric functions of θ with the given constraint. Function Value sec θ = -2 Constraint sin θ < 0

Summary:

The values of the six trigonometric functions of θ with the given constraints  sec θ = -2 and  sin θ < 0 are Sin θ = -√3/2, Cos θ = -1/2, Tan θ = √3, Cosec θ = -2/√3, Sec θ =  2/-1 = -2, Cot θ = 1/√3.