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$I f f (x) = \frac{\sin^{2} x + \cos^{4} x}{\cos^{2} x + \sin^{4} x}, t h e n f (2020) =$

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Correct option is B

f(x)=1−sin2⁡xcos2⁡x1−sin2⁡xcos2⁡x=1⇒f(2020)=1

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2023 | www.infinitylearn.com

If f(x)=sin2x+cos4xcos2x+sin4x, then f2020=

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$I f f (x) = \frac{\sin^{2} x + \cos^{4} x}{\cos^{2} x + \sin^{4} x}, t h e n f (2020) =$