1989 AHSME Problems/Problem 14

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 1

We have $\cot 10 +\tan 5=\frac{\cos 10}{\sin 10}+\frac{\sin 5}{\cos 5}=\frac{\cos10\cos5+\sin10\sin5}{\sin10\cos 5}=\frac{\cos(10-5)}{\sin10\cos5}=\frac{\cos5}{\sin10\cos5}=\boxed{(\textbf{B})\csc 10}$

Solution 2 (Half Angle Formulas)

$\cot 10 + \tan 5 = \frac{\cos 10}{\sin 10} + \frac{1 - \cos 10}{\sin 10} = \frac{1}{\sin 10} = \boxed{(\textbf{B})\csc 10}$ .

-j4andrews

Solution 3 (Double Angle Formulas)

$\cot 10 + \tan 5 = \frac{1}{\tan 10} + \tan 5 = \frac{1}{\frac{2 \tan 5}{1 - \tan^2 5}} + \tan 5 = \frac{1 - \tan^2 5}{2 \tan 5} + \tan 5 = \frac{1 - \tan^2 5}{2 \tan 5} + \frac{2 \tan^2 5}{2 \tan 5}$

$= \frac{1 + \tan^2 5}{2 \tan 5} = \frac{\sec^2 5}{2 \tan 5} = \frac{1}{2} \sec^2 5 \cot 5 = \frac{1}{2} \cdot \frac{1}{\cos^2 5} \cdot \frac {\cos 5}{\sin 5} = \frac{1}{2 \sin 5 \cos 5} = \frac{1}{\sin 10} = \boxed{(\textbf{B})\csc 10}$ .

-j314andrews

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
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These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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1989 AHSME Problems/Problem 14

Contents

Problem

Solution 1

Solution 2 (Half Angle Formulas)

Solution 3 (Double Angle Formulas)

See also