1989 AMC 12 Problem 14

Attempt Problem 14 of the 1989 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1989 AMC 12 solutions, or check the answer key.

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14.

\cot10+\tan5=

\csc5

\csc10

\sec5

\sec10

\sin15

Rewrite both terms using sine and cosine

Use

\sin10=2\sin5\cos5

and

\cos10=\cos^25-\sin^25

Using

\sin10=2\sin5\cos5,

\begin{aligned} \cot10+\tan5 &=\frac{\cos10}{\sin10}\\ &\quad+\frac{\sin5}{\cos5}. \end{aligned}

Combining the fractions gives numerator

\cos10+2\sin^25.

Since

\cos10=\cos^25-\sin^25,

this numerator is

1,

and the expression is

\csc10.

Thus the correct answer is B.

Problem 14 in other years