Olympiad Maths Prep

Track / Stage 4 / 194 of 340 #454 of 2000

Problem 454

AMC 12 late, AIME early
Algebra Difficulty 4.8 Find the answer

11 Given sin⁡(α+β)sin⁡(α−β)=3\frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}=3, then the value of tan⁡αtan⁡β\frac{\tan \alpha}{\tan \beta} is

Official solutions — 2

Solution 1

11 From the given, we have
sin⁡αcos⁡β+cos⁡αsin⁡β=3(sin⁡αcos⁡β−cos⁡αsin⁡β), \sin \alpha \cos \beta + \cos \alpha \sin \beta = 3(\sin \alpha \cos \beta - \cos \alpha \sin \beta),

which means
sin⁡αcos⁡β=2cos⁡αsin⁡β, \sin \alpha \cos \beta = 2 \cos \alpha \sin \beta,

so
tan⁡αtan⁡β=sin⁡αcos⁡βcos⁡αsin⁡β=2. \frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta} = 2.

Solution 2

sin⁡α⋅cos⁡β+cos⁡α⋅sin⁡β=3(sin⁡α⋅cos⁡β−cos⁡α⋅sin⁡β),sin⁡α⋅cos⁡β=2cos⁡α⋅sin⁡β.Therefore, tan⁡αtan⁡β=sin⁡α⋅cos⁡βcos⁡α⋅sin⁡β=2. \begin{array}{l} \sin \alpha \cdot \cos \beta + \cos \alpha \cdot \sin \beta \\ = 3(\sin \alpha \cdot \cos \beta - \cos \alpha \cdot \sin \beta), \\ \sin \alpha \cdot \cos \beta = 2 \cos \alpha \cdot \sin \beta. \\ \text{Therefore, } \frac{\tan \alpha}{\tan \beta} = \frac{\sin \alpha \cdot \cos \beta}{\cos \alpha \cdot \sin \beta} = 2. \end{array}

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.