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sin^4A/sin^2B + cos^4A/cos^2B = 1

to prove :

sin^4B/sin^2A + cos^4B/cos^2A = 1

$Given \frac{\sin^{4} A}{\sin^{2} B} + \frac{\cos^{4} A}{\cos^{2} B} = 1 \Rightarrow \frac{\sin^{4} A \cos^{2} B + \cos^{4} A \sin^{2} B}{\sin^{2} B \cos^{2} B} = 1 \Rightarrow \sin^{4} A (1 - \sin^{2} B) + (1 - \sin^{2} A)^{2} \sin^{2} B = \sin^{2} B \cos^{2} B \Rightarrow \sin^{4} A - \sin^{4} A \sin^{2} B + (1 + \sin^{4} A - 2 \sin^{2} A) \sin^{2} B = \sin^{2} B (1 - \sin^{2} B) \Rightarrow \sin^{4} A - \sin^{4} A \sin^{2} B + \sin^{2} B + \sin^{4} A \sin^{2} B - 2 \sin^{2} A \sin^{2} B = \sin^{2} B - \sin^{4} B \Rightarrow \sin^{4} A + \sin^{4} B - 2 \sin^{2} A \sin^{2} B = 0 \Rightarrow (\sin^{2} A - \sin^{2} B)^{2} = 0 \Rightarrow \sin^{2} A - \sin^{2} B = 0 \Rightarrow \sin^{2} A = \sin^{2} B \Rightarrow 1 - \cos^{2} A = 1 - \cos^{2} B \Rightarrow \cos^{2} A = \cos^{2} B Given \frac{\sin^{4} B}{\sin^{2} A} + \frac{\cos^{4} B}{\cos^{2} A} = \frac{\sin^{4} B}{\sin^{2} B} + \frac{\cos^{4} B}{\cos^{2} B} = \sin^{2} B + \cos^{2} B = 1$

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