1989 AHSME Problems/Problem 14

Revision as of 15:11, 25 February 2022 by Hastapasta (talk | contribs)
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Problem

$\cot 10+\tan 5=$


$\mathrm{(A) \csc 5 } \qquad \mathrm{(B) \csc 10 } \qquad \mathrm{(C) \sec 5 } \qquad \mathrm{(D) \sec 10 } \qquad \mathrm{(E) \sin 15 }$

Solution

We have \[\cot 10 +\tan 5=\frac{\cos 10}{\sin 10}+\frac{\sin 5}{\cos 5}=\frac{\cos10\cos5+\sin10\sin5}{\sin10\cos 5}=\frac{\cos(10-5)}{\sin10\cos5}=\frac{\cos5}{\sin10\cos5}=\csc10\]

See also

1989 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png