Difference between revisions of "1989 AHSME Problems/Problem 14"
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| + | == Problem == | ||
| + | |||
<math>\cot 10+\tan 5=</math> | <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 } </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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| + | == Solution 1 == | ||
| + | |||
| + | 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}=\boxed{(\textbf{B})\csc 10}</cmath> | ||
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| + | == Solution 2 (Half Angle Formulas) == | ||
| + | <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>. | ||
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| + | -j4andrews | ||
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| + | == Solution 3 (Double Angle Formulas) == | ||
| + | |||
| + | <math>\cot 10 + \tan 5 = \frac{1}{\tan 10} + \tan 5 = \frac{1}{\frac{2 \tan 5}{1 - \tan^2 5}} + \tan 5 = \frac{1 - \tan^2 5}{2 \tan 5} + \tan 5 = \frac{1 - \tan^2 5}{2 \tan 5} + \frac{2 \tan^2 5}{2 \tan 5} </math> | ||
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| + | <math>= \frac{1 + \tan^2 5}{2 \tan 5} = \frac{\sec^2 5}{2 \tan 5} = \frac{1}{2} \sec^2 5 \cot 5 = \frac{1}{2} \cdot \frac{1}{\cos^2 5} \cdot \frac {\cos 5}{\sin 5} = \frac{1}{2 \sin 5 \cos 5} = \frac{1}{\sin 10} = \boxed{(\textbf{B})\csc 10}</math>. | ||
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| + | -j314andrews | ||
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| + | == See also == | ||
| + | {{AHSME box|year=1989|num-b=13|num-a=15}} | ||
| + | |||
| + | [[Category: Introductory Trigonometry Problems]] | ||
| + | {{MAA Notice}} | ||
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. 
