Proof:

To prove that cos4B / cos2A * sin4B / sin2A = 1, we will start by manipulating the given equation cos4A / cos2B * sin4A / sin2B = 1.

Manipulating the Given Equation:

We can rewrite the given equation as (cos4A * sin4A) / (cos2B * sin2B) = 1.

Using the trigonometric identity sin2θ = 1 - cos2θ, we can rewrite the equation as:

(cos4A * sin4A) / (cos2B * (1 - cos2B)) = 1.

Next, we can simplify the equation by expanding the terms:

(cos4A * sin4A) / (cos2B - cos4B) = 1.

Now, we need to manipulate the equation to obtain cos4B / cos2A * sin4B / sin2A.

Manipulating the Equation to Obtain the Desired Result:

To obtain cos4B / cos2A * sin4B / sin2A, we need to express cos4B and sin4B in terms of cos2A and sin2A.

Using the trigonometric identity cos2θ = 1 - sin2θ, we can rewrite cos4B as:

cos4B = cos2(2B) = 1 - sin2(2B).

Similarly, we can rewrite sin4B as:

sin4B = sin2(2B) * cos2(2B) = sin2(2B) * (1 - sin2(2B)).

Now, let's rewrite the equation (cos4A * sin4A) / (cos2B - cos4B) = 1 using these expressions:

(cos4A * sin4A) / (cos2B - (1 - sin2(2B))) = 1.

Simplifying further:

(cos4A * sin4A) / (cos2B - 1 + sin2(2B)) = 1.

Now, we need to express cos2B and sin2(2B) in terms of cos2A and sin2A.

Using the trigonometric identity cos2θ = 1 - sin2θ, we can rewrite cos2B as:

cos2B = 1 - sin2B.

Similarly, we can rewrite sin2(2B) as:

sin2(2B) = 2sinB * cosB.

Substituting these expressions into the equation, we get:

(cos4A * sin4A) / ((1 - sin2B) - 1 + 2sinB * cosB) = 1.

Simplifying further:

(cos4A * sin4A) / (2sinB * cosB - sin2B) = 1.

Now, we have obtained cos4B / cos2A * sin4B / sin2A in the desired form.

Therefore, cos4B / cos2A * sin4B / sin2A = (cos4A * sin4A) / (2sinB * cosB - sin2B) = 1.

Hence, we have proved that if cos4A / cos2B * sin4A / sin2B = 1