Difference between revisions of "1989 AHSME Problems/Problem 14"
(Created page with "<math>\cot 10+\tan 5=</math> <math> \mathrm{(A) \csc 5 } \qquad \mathrm{(B) \csc 10 } \qquad \mathrm{(C) \sec 5 } \qquad \mathrm{(D) \sec 10 } \qquad \mathrm{(E) \sin 15 } </m...") |
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<math> \mathrm{(A) \csc 5 } \qquad \mathrm{(B) \csc 10 } \qquad \mathrm{(C) \sec 5 } \qquad \mathrm{(D) \sec 10 } \qquad \mathrm{(E) \sin 15 } </math> | <math> \mathrm{(A) \csc 5 } \qquad \mathrm{(B) \csc 10 } \qquad \mathrm{(C) \sec 5 } \qquad \mathrm{(D) \sec 10 } \qquad \mathrm{(E) \sin 15 } </math> | ||
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| + | 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> | ||
Revision as of 22:28, 29 February 2012
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)
