Since cot θ = Adjacent over hypotenuse and this is in the first quadrant, your triangle would look something like so:
................/
.............../.|
............../..|
.........../.....|
√(a²+1)/......|
......./.........|..1
...../...........|
../..............|
/θ_________
............a
Just ignore the dots, they're just place holders. So since cot = adjacent over hypotenuse, or cot = a/1 that's how the two sides would equal 1. Using the Pythagorean theorem, I found the hypotenuse of √(a² + 1).
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Verified answer
Since cot θ = Adjacent over hypotenuse and this is in the first quadrant, your triangle would look something like so:
................/
.............../.|
............../..|
.........../.....|
√(a²+1)/......|
......./.........|..1
...../...........|
../..............|
/θ_________
............a
Just ignore the dots, they're just place holders. So since cot = adjacent over hypotenuse, or cot = a/1 that's how the two sides would equal 1. Using the Pythagorean theorem, I found the hypotenuse of √(a² + 1).
So now for all the other trig values of theta:
sin θ = 1 / √(a² + 1)
cos θ = a / √(a² + 1)
tan θ = 1 / a
csc θ = √(a² + 1)
sec θ = √(a² + 1) / a
Hope it helps