Question:
If $f(x)=\frac{\sin ^{4} x+\cos ^{2} x}{\sin ^{2} x+\cos ^{4} x}$ for $x \in \mathrm{R}$, then $f(2002)=$
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(a) 1
Given:
$f(x)=\frac{\sin ^{4} x+\cos ^{2} x}{\sin ^{2} x+\cos ^{4} x}$
On dividing the numerator and denominator by $\cos ^{4}$
$f(x)=\frac{\tan ^{4} x+\sec ^{2} x}{1+\tan ^{2} x \sec ^{2} x}=\frac{1+\tan ^{4} x+\tan ^{2} x}{1+\tan ^{2} x\left(1+\tan ^{2} x\right)}=\frac{1+\tan ^{4} x+\tan ^{2} x}{1+\tan ^{4} x+\tan ^{2} x}=1$ (For every $x \in \mathrm{R}$ )
For x = 2002, we have
f (2002) = 1