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

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

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$1$

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(A) Given $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$.We know that $\sin^2 x + \cos^2 x = 1$,so $\cos^2 x = 1 - \sin^2 x$ and $\sin^2 x = 1 - \cos^2 x$.Substituting these into the numerator and denominator:Numerator: $\cos^2 x + \sin^4 x = \cos^2 x + \sin^2 x \cdot \sin^2 x = \cos^2 x + \sin^2 x(1 - \cos^2 x) = \cos^2 x + \sin^2 x - \sin^2 x \cos^2 x = 1 - \sin^2 x \cos^2 x$.Denominator: $\sin^2 x + \cos^4 x = \sin^2 x + \cos^2 x \cdot \cos^2 x = \sin^2 x + \cos^2 x(1 - \sin^2 x) = \sin^2 x + \cos^2 x - \cos^2 x \sin^2 x = 1 - \cos^2 x \sin^2 x$.Thus,$f(x) = \frac{1 - \sin^2 x \cos^2 x}{1 - \cos^2 x \sin^2 x} = 1$.Since $f(x) = 1$ for all $x \in R$,it follows that $f(2023) = 1$.

If $f(x) = \frac{\cos^2 x + \sin^4 x}{\sin^2 x + \cos^4 x}$ for all $x \in R$,then $f(2023) = $

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