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

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(A) Given the function $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.We know that $\sin^2 x = 1 - \cos^2 x$ and $\cos^2 x = 1 - \sin^2 x$.S

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(A) Given the function $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.We know that $\sin^2 x = 1 - \cos^2 x$ and $\cos^2 x = 1 - \sin^2 x$.Substituting these into the numerator and denominator:$f(x) = \frac{\cos^2 x + \sin^2 x (\sin^2 x)}{\sin^2 x + \cos^2 x (\cos^2 x)}$$f(x) = \frac{\cos^2 x + \sin^2 x (1 - \cos^2 x)}{\sin^2 x + \cos^2 x (1 - \sin^2 x)}$$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 $\sin^2 x + \cos^2 x = 1$, we get:$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \cos^2 x \sin^2 x} = 1$.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 $f(2002)$ is equal to

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