Question

If $f\left(x\right)=\frac{{\cos }^{2}x+{\sin }^{4}x}{{\sin }^{2}x+{\cos }^{4}x}$ for $x\in R$ then $f\left(2018\right)=$

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

$f\left(x\right)=\frac{co{s}^{2}x+si{n}^{4}x}{si{n}^{2}x+co{s}^{4}x}$

Now $co{s}^{2}x+si{n}^{4}x=co{s}^{2}x+(si{n}^{2}x{)}^2$

$=co{s}^{2}x+(1-co{s}^{2}x{)}^2$

$=co{s}^{2}x+1+co{s}^{4}x-2co{s}^{2}x$

$=co{s}^{4}x+1-co{s}^{2}x$

$=co{s}^{4}x+si{n}^{2}x$

$=$ Denominator

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

$\therefore f\left(2018\right)=1$