Maths Olympiad Prep

Track / Stage 3 / 129 of 260 #129 of 1964

Problem 129

AMC 10/12, early questions
Algebra Difficulty 3.3 Find the answer

If tan⁡x+tan⁡y=25\tan x+\tan y=25 and cot⁡x+cot⁡y=30\cot x + \cot y=30, what is tan⁡(x+y)\tan(x+y)?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Since cot⁡\cot is the reciprocal function of tan⁡\tan:
cot⁡x+cot⁡y=1tan⁡x+1tan⁡y=tan⁡x+tan⁡ytan⁡x⋅tan⁡y=30\cot x + \cot y = \frac{1}{\tan x} + \frac{1}{\tan y} = \frac{\tan x + \tan y}{\tan x \cdot \tan y} = 30
Thus, tan⁡x⋅tan⁡y=tan⁡x+tan⁡y30=2530=56\tan x \cdot \tan y = \frac{\tan x + \tan y}{30} = \frac{25}{30} = \frac{5}{6}
Using the tangent addition formula:
tan⁡(x+y)=tan⁡x+tan⁡y1−tan⁡x⋅tan⁡y=251−56=150\tan(x+y) = \frac{\tan x + \tan y}{1-\tan x \cdot \tan y} = \frac{25}{1-\frac{5}{6}} = \boxed{150}.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.