One other way to see it is to know that sin^4 -cos^4= sin^2-cos^2. Then everything cancels out and you get the result.

I did f(x) = (cos²x + sin²x.sin²x)/(sin²x + cos²x.cos²x) = (cos²x + sin²x.(1 - cos²x))/(sin²x + cos²x.(1 - sin²x)) = (cos²x + sin²x - sin²xcos²x))/(sin²x + cos²x - cos²xsin²x)) = (1 - sin²xcos²x)/(1 - sin²xcos²x) = 1. Very easy.

@matholympiadonlinesolver

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Nice

writing z for sin^2 ( x) one gets f ( x) = (1 - z + z^2) /( z + ( 1 - z) ^2) = (1 - z + z^2) /( 1 - z + z^2) = 1

@matholympiadonlinesolver

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Nice

How do you pronounce x?

@matholympiadonlinesolver

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What are you hearing?

Bro.. Nice explanation

@matholympiadonlinesolver

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Thank you

derivative of f(x) is 0. f(x) is constant and f(0)=1. hence f (x)=1 for all x.

@matholympiadonlinesolver

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Ok thank you

@davidsousaRJ

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@jwkim4428 how did you get thsi result? It is very tedious to calculate this derivative by the quotient rule, and there is no obvious term cancellation in the final result.