Question

If {(Cos^4)A/(Cos^2)B} + {(Sin^4)A/(Sin^2)B} = 1
Then prove that {(cos^4)B/(cos^2)A} + {(SIn^4)B/(sin^2)A} = 1

Answer 1

Answer:

Here is the explanation

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Answer 2

Cos⁴B/Cos²A  + Sin⁴B/Sin²A   = 1  if Cos⁴A/Cos²B  + Sin⁴A/Sin²B  = 1

Step-by-step explanation:

Cos⁴A/Cos²B  + Sin⁴A/Sin²B  = 1

=> Cos⁴ASin²B  + Sin⁴ACos²B  = Cos²BSin²B

=> Cos⁴A(1 - Cos²B)  + (Sin²A)²Cos²B  = Cos²BSin²B

=> Cos⁴A(1 - Cos²B)  + (1 - Cos²A)²Cos²B  = Cos²BSin²B

=> Cos⁴A(1 - Cos²B)  + (1 + Cos⁴A - 2Cos²A)Cos²B  = Cos²BSin²B

=> Cos⁴A - Cos⁴ACos²B  + Cos²B + Cos⁴ACos²B - 2Cos²ACos²B  = Cos²BSin²B

=> Cos⁴A  + Cos²B   - 2Cos²ACos²B  = Cos²B(1 - Cos²B)

=> Cos⁴A  + Cos²B   - 2Cos²ACos²B  = Cos²B - Cos⁴B

=>  Cos⁴A     - 2Cos²ACos²B  + Cos⁴B = 0

=> (cos²A - Cos²B)² = 0

=> cos²A - Cos²B = 0

=>  cos²A  =  Cos²B

=> 1 - Sin²A = 1 - Sin²B

=> Sin²A =  Sin²B

LHS = Cos⁴B/Cos²A  + Sin⁴B/Sin²A  

= Cos⁴B/Cos²B  + Sin⁴B/Sin²B

= Cos²B  + Sin²B

= 1

= RHS

QED

Proved

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