If f(x) = frac{cos^2 x + sin^4 x}{sin^2 x + c | vedclass

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(A) Given $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.Using the identity $\sin^4 x = \sin^2 x (1 - \cos^2 x)$ and $\cos^4 x = \cos^2 x (1

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$1$

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(A) Given $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.Using the identity $\sin^4 x = \sin^2 x (1 - \cos^2 x)$ and $\cos^4 x = \cos^2 x (1 - \sin^2 x)$,we substitute these into the expression:$f(x) = \frac{\cos^2 x + \sin^2 x (1 - \cos^2 x)}{\sin^2 x + \cos^2 x (1 - \sin^2 x)}$Expanding the terms:$f(x) = \frac{\cos^2 x + \sin^2 x - \sin^2 x \cos^2 x}{\sin^2 x + \cos^2 x - \cos^2 x \sin^2 x}$Since $\sin^2 x + \cos^2 x = 1$,the numerator and denominator become identical:$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \sin^2 x \cos^2 x} = 1$Therefore,for any $x \in R$,$f(x) = 1$. Thus,$f(2002) = 1$.

If $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$ for $x \in R$,then $f(2002) = $

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