Answer:
sinθ = -5/√61
secθ = √61/6
tanθ = -5/6
Step-by-step explanation:
From the given coordinate (6, -5), x = 6 and y = -5. This shows that the point lies in the 4th quadrant. In the fourth quadrant, only cos θ is positive, both sin θ and tan θ are negative.
Let us get the value of the radius 'r' first before calculating the trigonometry identities.
Using the Pythagoras theorem;
![r^2 = x^2 + y^2\\r^2 = 6^2 + (-5)^2\\r^2 = 36+25\\r^2 = 61\\r = \sqrt{61}](https://tex.z-dn.net/?f=r%5E2%20%3D%20x%5E2%20%2B%20y%5E2%5C%5Cr%5E2%20%3D%206%5E2%20%2B%20%28-5%29%5E2%5C%5Cr%5E2%20%3D%2036%2B25%5C%5Cr%5E2%20%3D%2061%5C%5Cr%20%3D%20%5Csqrt%7B61%7D)
Using SOH, CAH, TOA to get the trigonometry identities;
Given x = 6, y =5 and r = √61
![sin \theta = opp/hyp = y/r\\sin \theta = 5/\sqrt{61} \\](https://tex.z-dn.net/?f=sin%20%5Ctheta%20%3D%20opp%2Fhyp%20%3D%20y%2Fr%5C%5Csin%20%5Ctheta%20%3D%205%2F%5Csqrt%7B61%7D%20%5C%5C)
Since sin θ, is negative in the fourth quadrant, ![sin \theta = -5/\sqrt{61} \\](https://tex.z-dn.net/?f=sin%20%5Ctheta%20%3D%20-5%2F%5Csqrt%7B61%7D%20%5C%5C)
![cos \theta = adj/hyp = x/r\\cos \theta = 6/\sqrt{61} \\](https://tex.z-dn.net/?f=cos%20%5Ctheta%20%3D%20adj%2Fhyp%20%3D%20x%2Fr%5C%5Ccos%20%5Ctheta%20%3D%206%2F%5Csqrt%7B61%7D%20%5C%5C)
![sec \theta = \frac{1}{cos \theta} \\sec \theta = \frac{1}{6/\sqrt{61} } \\sec \theta = \frac{\sqrt{61} }{6}](https://tex.z-dn.net/?f=sec%20%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7Bcos%20%5Ctheta%7D%20%5C%5Csec%20%5Ctheta%20%3D%20%5Cfrac%7B1%7D%7B6%2F%5Csqrt%7B61%7D%20%20%7D%20%5C%5Csec%20%5Ctheta%20%3D%20%5Cfrac%7B%5Csqrt%7B61%7D%20%7D%7B6%7D)
For tanθ:
![tan \theta = \frac{opp}{adj} = \frac{y}{x} \\tan \theta = \frac{5}{6}\\](https://tex.z-dn.net/?f=tan%20%5Ctheta%20%3D%20%5Cfrac%7Bopp%7D%7Badj%7D%20%3D%20%5Cfrac%7By%7D%7Bx%7D%20%20%5C%5Ctan%20%5Ctheta%20%3D%20%5Cfrac%7B5%7D%7B6%7D%5C%5C)
Since tan θ, is negative in the fourth quadrant, ![tan \theta = -5/6](https://tex.z-dn.net/?f=tan%20%5Ctheta%20%3D%20-5%2F6)