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If $f(x)=\frac{\cos x}{\sin ^2 x+\cos ^4 x}$, for $x \in R$, then $f(2002)$ is equal to
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Verified Answer
The correct answer is:
$1$
We have,
$$
f(x)=\frac{\cos ^2 x+\sin ^4 x}{\sin ^2 x+\cos ^4 x}
$$
$\begin{aligned} & =\frac{1-\sin ^2 x+\sin ^4 x}{1-\cos ^2 x+\cos ^4 x} \\ & =\frac{1-\sin ^2 x\left(1-\sin ^2 x\right)}{1-\cos ^2 x\left(1-\cos ^2 x\right)} \\ & =\frac{1-\sin ^2 x \cos ^2 x}{1-\cos ^2 x \sin ^2 x}=1 \\ \therefore \quad f(2002) & =1\end{aligned}$
$$
f(x)=\frac{\cos ^2 x+\sin ^4 x}{\sin ^2 x+\cos ^4 x}
$$
$\begin{aligned} & =\frac{1-\sin ^2 x+\sin ^4 x}{1-\cos ^2 x+\cos ^4 x} \\ & =\frac{1-\sin ^2 x\left(1-\sin ^2 x\right)}{1-\cos ^2 x\left(1-\cos ^2 x\right)} \\ & =\frac{1-\sin ^2 x \cos ^2 x}{1-\cos ^2 x \sin ^2 x}=1 \\ \therefore \quad f(2002) & =1\end{aligned}$
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