tant = sint/cost

but also

tanp = cost + sint / cost - sint

p = t + pi/4

basically, think of it as rotating the entire frame by pi/4 and then doing stuff on this plane as we usually do with unit circle trigonometry. BUT, in this new plane, still calculating things wrt old frame (t co-ordinates), and doing that gives us cost + sint / cost - sint.

cost + sint and cost - sint both have range -√2 to √2. i.e. they parameterize the circle u^2 + v^2 = 2,

cost + sint = √2 sin(t + pi/4) = √2sinp = u

cost - sint = √2 cos(t + pi/4) = √2cosp = v

Suppose, we only restrict our original domain t to -pi/4 to pi/4. why this ?? because this is enough to make them go around the first quarter.

cost + sint = √2sint(t + pi/4) , suppose we start from angle -pi/4 and move till angle 0, that'd make our point move from 0 to 1. and as we move further pi/4 from 0, then we move from 1 to √2. But this is same as moving on circle of radius √2 angle 0 to pi/2.

A(t) = arcsin(√2sint) = arcsin(p)

B(t) = arcosh(√2cost) = arcosh(q)

arcsin naturally reverses the relationship to link p to the standard way to relate a vertical coordinate to an angle using the unit circle. Here we're linking the vertical co-ordinate of a point on the √2 radius circle to the angle on the unit circle (arc length), we don't know what this angle is (it's not p or t obv), but there is obviously one angle A(t) such that sin(A(t)) = √2sint. You might think this breaks trigonometry, because LHS is between -1 and RHS is -√2 and √2, but no, remember, i said, t only goes from -pi/4 to pi/4., so, RHS too stays between -1 to 1.

if we take arcos here to reverse the relation to link q to the horizontal co-ordinate to angle on unit circle,

arccos(q) = arcos(√2cost), then we immediately come across a problem. we have defined t for only values from -pi/4 to pi/4.

when t goes from -pi/4 to 0, √2cost goes from 1 to √2, and when we move further pi/4, we go from √2 to 1. This goes beyond 1. And that is not allowed.

cos(c(t)) = √2cost is > 1,

such c(t) doesn't exist unless we allow imaginary inputs. So yeah, that's literally an okay solution, but i want to avoid imaginary numbers for now.

instead of reversing the relation to link q to the horizontal co-ordinate to the angle on unit circle, we link it to the angle on the unit hyperbola., why unit hyperbola and nothing else ? because, the x co-ordinates of points on the unit hyperbola are always greater than or equal to 1 and for hyperbolic trig, if our angle is between -pi/4 and pi/4, it doesn't even exceeds √2. That's what we want.

As our angle goes from -pi/4 to 0, the x co-ordinates of the point on the unit hyperbola change from 1 to √2 and again from √2 to 1.

H(t) = arccosh(√2cost) = arccosh(q)

Now, one of the coolest thing about this transformation is that if we take a small differential of this circular angle or hyperbolic angle., then what we get is purely in terms of pure trigonometry (the unit circle trigonometry we're familiar with). It's rare, but it happens here, the reason why this happens is because of the way we chose our functions to reverse the relationship and link back the co-ordinates to the angle and our precise choice of choosing +45 degree rotation frame, what this does is, changing the co-ordinates by changing angle dt corresponds to changing angle dA in a specified and fix manner, like how in integral of secx proof, I showed dp = secxdx, and this angle allows the points to move smoothly. Like for unit hyperbola, we have dp = secxdx, this is a small change in hyperbolic angle, and it has a trigonometric term in it. This quantity, dA, dH and thus √tanxdx measures that non-euclidean factor.

And what next ? I just added and subtracted them, because I saw cost and sint. Subtracting them gives the integral of square root of tanx and adding them gives the integral of square root cotx.

1) the hyperbola that i drew should be x^2 - y^2 = 1, not x^2 - y^2 = 2. Because i parameterized it with cosh(H) and sinh(H). But that doesn't change anything. All the ideas remains same.

2) On the first page top right side, i wrote d(cosx) = sinxdx. It should be -sinxdx. But again, I have never used this anywhere later in the proof, so doesn't make any difference.