Answer:
Step-by-step explanation:
<u>Given function:</u>
<u>Plot two points and check which graph has those:</u>
and
Points (0, 2) and (1, 6) are correct on the image 1 with blue graph
Answer:
and ![\sec \theta = -\dfrac{13}{12}](https://tex.z-dn.net/?f=%5Csec%20%5Ctheta%20%3D%20-%5Cdfrac%7B13%7D%7B12%7D)
Step-by-step explanation:
Assume that the terminal side of thetaθ passes through the point (−12,5).
In ordered pair (-12,5), x-intercept is negative and y-intercept is positive. It means the point lies in 2nd quadrant.
Using Pythagoras theorem:
![hypotenuse^2=perpendicular^2+base^2](https://tex.z-dn.net/?f=hypotenuse%5E2%3Dperpendicular%5E2%2Bbase%5E2)
![hypotenuse^2=(5)^2+(12)^2](https://tex.z-dn.net/?f=hypotenuse%5E2%3D%285%29%5E2%2B%2812%29%5E2)
![hypotenuse^2=25+144](https://tex.z-dn.net/?f=hypotenuse%5E2%3D25%2B144)
![hypotenuse^2=169](https://tex.z-dn.net/?f=hypotenuse%5E2%3D169)
Taking square root on both sides.
![hypotenuse=13](https://tex.z-dn.net/?f=hypotenuse%3D13)
In a right angled triangle
![\sin \theta = \dfrac{opposite}{hypotenuse}](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%20%3D%20%5Cdfrac%7Bopposite%7D%7Bhypotenuse%7D)
![\sin \theta = \dfrac{5}{13}](https://tex.z-dn.net/?f=%5Csin%20%5Ctheta%20%3D%20%5Cdfrac%7B5%7D%7B13%7D)
![\sec \theta = \dfrac{hypotenuse}{adjacent}](https://tex.z-dn.net/?f=%5Csec%20%5Ctheta%20%3D%20%5Cdfrac%7Bhypotenuse%7D%7Badjacent%7D)
![\sec \theta = \dfrac{13}{12}](https://tex.z-dn.net/?f=%5Csec%20%5Ctheta%20%3D%20%5Cdfrac%7B13%7D%7B12%7D)
In second quadrant only sine and cosecant are positive.
and ![\sec \theta = -\dfrac{13}{12}](https://tex.z-dn.net/?f=%5Csec%20%5Ctheta%20%3D%20-%5Cdfrac%7B13%7D%7B12%7D)
Answer: (0,2), (3,0), (6, -2)
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