Problem:
Suppose that secx+tanx=722β and that cscx+cotx=nmβ, where nmβ is in lowest terms. Find m+n.
Solution:
Since sec2xβtan2x=1, we see that secxβtanx=1/p, where p stands for 22/7. This leads to
2secx=p+p1β and 2tanx=pβp1β,
and then to
cosx=p2+12pβ and sinx=p2+1p2β1β.
It is now an easy matter to show that
cscx+cotx=pβ1p+1β=1529β.
Thus, m+n=44β.
OR
We apply the half-angle identities
tan(x/2)=cscxβcotx and cot(x/2)=cscx+cotx.
Since
722β=secx+tanx=csc(2Οβ+x)βcot(2Οβ+x)=tan(21β(2Οβ+x))=tan(4Οβ+2xβ)
we have
nmβ=cscx+cotx=cot2xββ=tan(2Οββ2xβ)=tan(43Οββ(4Οβ+2xβ))=tan(43Οββarctan722β)=1β722ββ1β722ββ=1529β.β
The problems on this page are the property of the MAA's American Mathematics Competitions