已知 $\sin \alpha + \cos \beta = \dfrac{{\sqrt 3 }}{2}$，$\cos \alpha + \sin \beta = \sqrt 2 $，求 $\tan \alpha \cdot \cot \beta $ 的值．
【难度】
【出处】
无
【标注】
【答案】
$- \dfrac{7}{{73}}$
【解析】
两式平方，有$${\sin ^2}\alpha + {\cos ^2}\beta + 2\sin \alpha \cos \beta = \dfrac{3}{4},$$$${\cos ^2}\alpha + {\sin ^2}\beta + 2\cos \alpha \sin \beta = 2.$$上述两式相加减，有$$2 + 2\sin \left( {\alpha + \beta } \right) = \dfrac{{11}}{4},$$$$2\sin \left( {\alpha + \beta } \right)\sin \left( {\alpha - \beta } \right) + 2\sin \left( {\alpha - \beta } \right) = - \dfrac{5}{4}.$$于是$$\sin \left( {\alpha + \beta } \right) = \dfrac{3}{8},$$$$\sin \left( {\alpha - \beta } \right) = - \dfrac{5}{{11}}.$$所以$$2\sin \alpha \cos \beta = \dfrac{3}{8} - \dfrac{5}{{11}} = - \dfrac{7}{{88}},$$$$2\cos \alpha \sin \beta = \dfrac{3}{8} + \dfrac{5}{{11}} = \dfrac{{73}}{{88}}.$$因此$$\tan \alpha \cdot \cot \beta = \dfrac{{2\sin \alpha \cos \beta }}{{2\cos \alpha \sin \beta }} = - \dfrac{7}{{73}}.$$
答案
解析
备注
