Difference between revisions of "1989 AHSME Problems/Problem 14"
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<math>\cot 10 + \tan 5 = \frac{\cos 10}{\sin 10} + \frac{1 - \cos 10}{\sin 10} = \frac{1}{\sin 10} = \boxed{(\textbf{B})\csc 10}</math>. | <math>\cot 10 + \tan 5 = \frac{\cos 10}{\sin 10} + \frac{1 - \cos 10}{\sin 10} = \frac{1}{\sin 10} = \boxed{(\textbf{B})\csc 10}</math>. | ||
| + | -j4andrews | ||
== Solution 3 (Double Angle Formulas) == | == Solution 3 (Double Angle Formulas) == | ||
Latest revision as of 01:45, 11 July 2025
Contents
Problem
Solution 1
We have ![\[\cot 10 +\tan 5=\frac{\cos 10}{\sin 10}+\frac{\sin 5}{\cos 5}=\frac{\cos10\cos5+\sin10\sin5}{\sin10\cos 5}=\frac{\cos(10-5)}{\sin10\cos5}=\frac{\cos5}{\sin10\cos5}=\boxed{(\textbf{B})\csc 10}\]](http://latex.artofproblemsolving.com/4/d/e/4de150c274c2995c8d47685b217c43d26a8b0b88.png)
Solution 2 (Half Angle Formulas)
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-j4andrews
Solution 3 (Double Angle Formulas)
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-j314andrews
See also
| 1989 AHSME (Problems • Answer Key • Resources) | ||
| 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. 
