tan alpha = 1/6, what is sec, cos, cot, sin, and csc?

Go online to this page and learn the relationship between the 6 Trig. functions:

http://www.efunda.com/math/trig_functions/trig_relation.cfm

tangent = opposite / adjacent

a² + b² = c²

6² + 1² = (hypotenuse)²

√(37) = hypotenuse

sec = hypotenuse / adjacent

sec = √(37) / 6 ≈ 1.014

cos = adjacent / hypotenuse

cos = 6/√(37)

cos = [6√(37)] / 37 ≈ 0.986

cot = adjacent / opposite

cot = 6 / 1 = 6

sin = opposite / hypotenuse

sin = 1 / √(37)

sin = √(37) / 37 ≈ 0.164

csc = hypotenuse / opposite

csc = √(37) / 1 = √(37) ≈ 6.083

If the tangent of an angle is 1/6, there are a few ways to find the other trig values.  First, cot(alpha) = 1/tan(alpha) so 1 over 1/6 is 6.  The formula sin²(alpha) + cos²(alpha)=1 can be divided through by cos² to give tan²(alpha) + 1= sec²(alpha).  The problem with plugging the 1/6 into that is you do not know which part of the circle or graph it is in so sec(alpha) upon solving would be sqrt[1/36 + 1] = sqrt[37/36] = sqrt(37)/6.  It might be negative depending on the original angle but let's just assume it is positive for now.  That brings us to sin and cos.   Since sec(alpha) = 1/cos(alpha), it = the reciprocal of sqrt(37)/6 which is 6*sqrt(37)/37.  Finally plug that back into sin²(alpha) + cos²(alpha)=1 to solve for cos(alpha).  

To summarize, use cot(alpha) = 1/tan(alpha) to get cot(alpha).  Use sec(alpha)= sqrt(tan²(alpha)+1) and then use that to get cos(alpha) since cos(alpha) = 1/sec(alpha).  Finally, use sin²(alpha) + cos²(alpha)=1 to get sin(alpha).  Since tan is positive in both quadrants 1 and 3, the sign on some of the trig functions might differ.