Question

if cos^4A/cos^2B + sin^4A/sin^2B=1. prove that sin^4A+sin^4B = 2sin^2A.sin^2B

Answer 1

simplify the statement first

Answer 2

Answer:

Step-by-step explanation:

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.