If tanθ=2, find the six trigonometric functions of θ.

The cotangent is easy, since
cot θ = 1/(tan θ) = 1/2

Next, find the secant, using the tan-sec version of Pythagoras:

tan² θ + 1 = sec² θ
sec ² θ = 2² + 1 = 5
sec θ = √5

Next sec θ = 1/(cos θ), so
cos θ = 1/(√5) = √5 / 5

"Regular" Pythagoras is cos² θ + sin² θ = 1, so:

sin θ = √(1 - cos² θ) = √(1 - 5/25) = √(20/25) = 2√5 / 5

Finally, the cosecant is:

csc θ = 1/(sin θ) = 5 / (2 √5) = √5 / 2

tan @ =2/1
sqrt (2^2 +1^1) =sqrt5
sin @ =2/sqrt5 =2sqrt5/5
cos@ =1/sqrt5 = sqrt5/5