Problem:
If f(x−1x)=x1 for all x=0,1 and 0<θ<2π, then f(sec2θ)=
Answer Choices:
A. sin2θ
B. cos2θ
C. tan2θ
D. cot2θ
E. csc2θ
Solution:
If x−1x=sec2θ then x=xsec2θ−sec2θ, or sec2θ=x(sec2θ−1)= xtan2θ. Hence x=sin2θ1 and f(sec2θ)=sin2θ.
OR
First solve y=x−1x for x to find x=y−1y. Then f(y)=yy−1. Hence
f(sec2θ)=sec2θsec2θ−1=1−cos2θ=sin2θ.
OR
Since f⎝⎜⎜⎛1−x11⎠⎟⎟⎞=f(x−1x)=x1,(∗)
f(sec2θ)=f(cos2θ1)=f(1−sin2θ1)=sin2θ
where for the last equality we substituted sin2θ for x1 in (∗).