Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Formula:- ${\sin }^{2}x+{\cos }^{2}x=1$

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

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

$f\left(2002\right)$$=\frac{1-{\sin }^2\left(2002\right){\cos }^2\left(2002\right)}{1-{\cos }^2\left(2002\right){\sin }^2\left(2002\right)}=1$