Find \( f(2002) \)
If
\[ f(x)=\frac{\sin^4x+\cos^2x}{\sin^2x+\cos^4x}, \qquad x\in R \]
then \( f(2002) \) is equal to
(a) \(1\)
(b) \(2\)
(c) \(3\)
(d) \(4\)
Let
\[ \sin^2x=a,\qquad \cos^2x=b \]
Then,
\[ a+b=1 \]
Now,
\[ f(x)=\frac{a^2+b}{a+b^2} \]
Since
\[ b=1-a \]
numerator:
\[ a^2+b = a^2+1-a \]
denominator:
\[ a+b^2 = a+(1-a)^2 = a+1-2a+a^2 = a^2-a+1 \]
Numerator = Denominator
Therefore,
\[ f(x)=1 \]
Hence,
\[ f(2002)=1 \]
\[ \boxed{\text{Correct Answer: (a)}} \]