We need to evaluate cos(arctan(x)).
We perform a substitution:
y = arctan(x)
tan(y) = x
Solution 1
We now need to find cos(y).
Now,
tan(y) = sin(y)/cos(y) = x/1
We imagine a triangle with angle y in one corner and angle 90. The side adjacent to angle y (cos y) is set to length 1, and the opposite (sin y) is set to x.
The hypotenuse of this triangle is:
x^2+1 = h^2
sqrt(x^2+1) = h
(sqrt is the square root function.)
cos(y) is the adjacent side of length 1 over the hypotenuse that we've just calculated.
cos(y) = 1/sqrt(x^2+1)
So:
cos(arctan(x)) = 1/sqrt(x^2+1)
Solution 2
We now need to find cos(y).
Now,
tan(y) = sin(y)/cos(y) = x
Can we eliminate sin(y) and just have cos(y)?
sin^2(y)/cos^2(y) = x^2
If we can add cos^2(y) to sin^2(y), it can be simplified.
sin^2(y)/cos^2(y) + cos^2(y)/cos^2(y) = x^2 + cos^2(y)/cos^2(y)
[sin^2(y) + cos^2(y)]/cos^2(y) = x^2 + 1
1/cos^2(y) = x^2+1 (because of the general identity that sin^2(a)+cos^2(a)=1).
1/(x^2+1) = cos^2(y)
1/sqrt(x^2+1) = cos(y)
cos(y) = 1/sqrt(x^2+1)
So:
cos(arctan(x)) = 1/sqrt(x^2+1)