Equations that are indented are NOT equal to the initial equation. They are just
manipulations of various parts of the initial equation (essentially getting secant
and cotangent in terms of cosine and sine).
sec^2(pi/2 - x) -1 = cot^2(x)
sec(x) = 1 / cos(x) so:
sec^2(pi/2 - x) = 1 / cos^2(pi/2 - x)
The given equation then becomes
[1 / cos^2(pi/2 - x)] - 1 = cot^2(x)
cot(x) = 1 / tan(x)
tan(x) = sin(x) / cos(x) so:
cot(x) = 1 / tan(x) = cos(x) / sin(x)
cot^2(x) = [cos(x) / sin(x)]^ = cos^2(x) / sin^2(x)
The equation becomes
[1 / cos^2(pi/2 - x)] - 1 = cos^2(x) / sin^2(x)
cos(pi/2 - x) = sin(x) so:
cos^2(pi/2 - x) = sin^2(x)
The equation becomes
[1 / sin^2(x)] - 1 = cos^2(x) / sin^2(x)
Multiply both sides by sin^2(x):
[ [1 / sin^2(x)] - 1 ]*sin^2(x) = [cos^2(x) / sin^2(x)]*sin^2(x)
1 - sin^2(x) = cos^2(x)
cos^2(x) + sin^2(x) = 1
This is a well-known trigonometric identity and should be enough to show
that the initial equation is valid, but I'll prove this identity for the sake of
completeness:
The Pythagorean Theorem: A^2 +B^2 = C^2
sine is the ratio of the side opposite the angle to the hypotenuse.
cosine is the ratio of the side adjacent to the angle and the hypotenuse.
If we take a triangle with the right angle at the bottom left corner and put
that corner at the origin on a graph, the side opposite the bottom right corner
is along the y axis and the side adjacent to that angle is along the x axis.
So we adjust our Pythagorean Theorem variables to: x^2 + y^2 = h^2
where h is the hypotenuse.
(x^2) / (h^2) + (y^2) / (h^2) = (h^2) / (h^2)
(x/h)^2 + (y/h)^2 = 1
ratio of x to h is: x/h = cos(angle)
ratio of y to h is: y/h = sin(angle)
plug back into equation (x/h)^2 + (y/h)^2 = 1
cos^2(angle) + sin^2(angle) = 1
