Question

If $f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{4}x}{{\sin }^{2}x+{\cos }^{4}x}\forall x\in \text{R}$ then show that $f\left(2012\right)=1$.

Answer 1

$f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{4}x}{{\sin }^{2}x+{\cos }^{4}x}$, show that $f\left(2012\right)=1$

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

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

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

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

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

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

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

$f\left(x\right)=1$

$\therefore f\left(2012\right)=1$.