Solution:

Given:
m = cos A - sin A
n = cos A sin A

We need to prove that:
√m/n √n/m = 2/√(1 - tan^2 A)

Proof:

To begin the proof, let's first simplify the expressions for m and n.

Simplifying m:
m = cos A - sin A

Using the trigonometric identity sin^2 A + cos^2 A = 1, we can rewrite sin A as √(1 - cos^2 A).

m = cos A - √(1 - cos^2 A)

Next, we can rationalize the denominator by multiplying both the numerator and denominator by the conjugate of the denominator, which is cos A + √(1 - cos^2 A).

m = (cos A - √(1 - cos^2 A))(cos A + √(1 - cos^2 A))
m = cos^2 A - (1 - cos^2 A)
m = 2cos^2 A - 1

Simplifying n:
n = cos A sin A

Using the trigonometric identity sin 2A = 2sin A cos A, we can rewrite n as sin 2A/2.

n = (sin 2A)/2

Calculating √m/n:
Now, let's calculate √m/n.

√m/n = √((2cos^2 A - 1)/(sin 2A/2))
√m/n = √((2cos^2 A - 1)/(sin A cos A))
√m/n = √((2cos^2 A - 1)/(2sin A cos A))
√m/n = √((cos^2 A - 1/2)/(sin A cos A))
√m/n = √((1 - 2sin^2 A)/(sin A cos A))
√m/n = √((1 - sin^2 A - sin^2 A)/(sin A cos A))
√m/n = √((cos^2 A)/(sin A cos A))
√m/n = √(cos A/sin A)
√m/n = √(1/tan A)
√m/n = 1/√(tan A)
√m/n = √(1/tan^2 A)

Calculating √n/m:
Now, let's calculate √n/m.

√n/m = √((sin 2A/2)/(2cos^2 A - 1))
√n/m = √((sin A cos A)/(2cos^2 A - 1))
√n/m = √((sin A cos A)/(cos^2 A - sin^2 A))
√n/m = √((sin A cos A)/(cos A + sin A)(cos A - sin A))
√n/m = √(1/(cos A + sin A))
√n/m = 1/√(cos A + sin A)

Proving √m/n √n/m = 2/√(1 - tan^2 A):