Difference between revisions of "1989 AHSME Problems/Problem 14"
| Line 5: | Line 5: | ||
We have <cmath>\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</cmath> | We have <cmath>\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</cmath> | ||
| + | {{MAA Notice}} | ||
Revision as of 13:48, 5 July 2013
We have ![\[\cot 10 +\tan 5=\frac{\cos 10}{\sin 10}+\frac{\sin 5}{\cos 5}=\frac{\cos10\cos5+\sin10\sin5}{\sin10\cos 5}=\frac{\cos(10-5)}{\sin10\cos5}=\frac{\cos5}{\sin10\cos5}=\csc10\]](http://latex.artofproblemsolving.com/f/a/e/fae4baa917f7824db4ba503601d46524564dafa7.png)
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
