Problem Contributed By Community Member debroy0611

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Problem Contributed By Community Member debroy0611

This is an adaptation of a relatively simple problem from the 2024 GaoKao.

Problem contributed by debroy0611

$\tan \alpha \tan \beta = 2$

$\Rightarrow \dfrac{\sin \alpha \sin \beta}{\cos \alpha \cos \beta} = 2$

$\Rightarrow \sin \alpha \sin \beta = 2 \cos \alpha \cos \beta$

$\cos(\alpha + \beta) = \dfrac{1}{4}$

$\Rightarrow \cos \alpha \cos \beta - \sin \alpha \sin \beta = \dfrac{1}{4}$

$\Rightarrow \cos \alpha \cos \beta - 2 \cos \alpha \cos \beta = \dfrac{1}{4}$

$\therefore \cos \alpha \cos \beta = - \dfrac{1}{4}$ and $\sin \alpha \sin \beta = -\dfrac{1}{2}$

$\therefore \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta = - \dfrac{3}{4} = -0.75$