Question

Cos^4A/cos^2B+sin^4A/sin^2B=1
Then prove
Cos^4B/cos^2A+sin^4B/sin2A=1

Answer 1

Starting with: 
1 = (cos(A))^4 / (cos(B))^2 + (sin(A))^4 / (sin(B))^2 

define: 
a = cos(A) 
b = cos(B) 
Notice that 
sin(A) = sqrt(1 - (cos(A))^2) = sqrt(1-a^2) , so: 
(sin(A))^2 = 1 - a^2 , and likewise: 
(sin(B))^2 = 1 - b^2 

Therefore, your original equation becomes: 
1 = a^4/b^2 + (1 - a^2)^2/(1 - b^2) 

Multiply both sides by (b^2) * (1 - b^2): 
(b^2) * (1 - b^2) = (a^4)*(1 - b^2) + (b^2)*(1 - a^2)^2 
b^2 - b^4 = a^4 - (a^4)*(b^2) + (b^2)*(1 - 2*a^2 + a^4) 
b^2 - b^4 = a^4 - (a^4)*(b^2) + b^2 - 2*(a^2)*(b^2) + (a^4)(b^2) 
Cancelling out some terms: 
- b^4 = a^4 - 2*(a^2)(b^2) , or 
0 = a^4 - 2*(a^2)(b^2) + b^4 
= (a^2 - b^2)^2 

=> a^2 = b^2 

OK, that means 
(cos(B))^4 = b^4 
(cos(A))^2 = a^2 = b^2 
so: 
(cos(B))^4 / (cos(A))^2 = b^4 / b^2 = b^2 

It also means that: 
(sin(B))^4 = ((sin(B))^2)^2 = (1 - b^2)^2 , and 
(sin(A))^2 = (1 - a^2) = (1 - b^2) 
so: 
(sin(B))^4 / (sin(A))^2 = (1 - b^2)^2 / (1 - b^2) = 1 - b^2 

So when you add the sum together, you get: 
(cos(B))^4 / (cos(A))^2 + (sin(B))^4 / (sin(A))^2 = b^2 + 1 - b^2 = 1 

Basically, the point is that the only way first equation could work is if (cos(A))^2 = (cos(B))^2 (and thus automatically for the sine functions as well); so if that's true, the second equation follows automatically.