Problem:
Suppose that secx+tanx=722 and that cscx+cotx=nm, where nm is in lowest terms. Find m+n.
Solution:
Since sec2x−tan2x=1, we see that secx−tanx=1/p, where p stands for 22/7. This leads to
2secx=p+p1 and 2tanx=p−p1,
and then to
cosx=p2+12p and sinx=p2+1p2−1.
It is now an easy matter to show that
cscx+cotx=p−1p+1=1529.
Thus, m+n=44.
OR
We apply the half-angle identities
tan(x/2)=cscx−cotx and cot(x/2)=cscx+cotx.
Since
722=secx+tanx=csc(2π+x)−cot(2π+x)=tan(21(2π+x))=tan(4π+2x)
we have
nm=cscx+cotx=cot2x=tan(2π−2x)=tan(43π−(4π+2x))=tan(43π−arctan722)=1−722−1−722=1529.
The problems on this page are the property of the MAA's American Mathematics Competitions