Question

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

Answer 1

$f\left(x\right)=\frac{({\sin }^{2}x{)}^2+{\cos }^{2}x}{1-{\cos }^{2}x+({\cos }^{2}x{)}^2}$

$=\frac{(1-{\cos }^{2}x{)}^2+{\cos }^{2}x}{1-{\cos }^{2}x+{\cos }^{4}x}$

$=\frac{1+{\cos }^{4}x-2{\cos }^{2}x+{\cos }^{2}x}{1-{\cos }^{2}x+{\cos }^{4}x}$

$=\frac{1+{\cos }^{4}x-{\cos }^{2}x}{1-{\cos }^{2}x+{\cos }^{4}x}=1$

$\text{Now},f\left(2002\right)=1$

Therefore, option (A) is correct.