Chứng minh
\(\dfrac{{\sin \alpha + \sin \beta \cos \left( {\alpha + \beta } \right)}}{{\cos \alpha - \sin \beta \sin \left( {\alpha + \beta } \right)}} = \tan \left( {\alpha + \beta } \right)\) (khi các biểu thức có nghĩa)
Advertisements (Quảng cáo)
\(\begin{array}{l}\dfrac{{\sin \alpha + \sin \beta \cos \left( {\alpha + \beta } \right)}}{{\cos \alpha - \sin \beta \sin \left( {\alpha + \beta } \right)}}\\ = \dfrac{{\sin \alpha + \dfrac{1}{2}\left[ {\sin \left( {\alpha + 2\beta } \right) - \sin \alpha } \right]}}{{\cos \alpha + \dfrac{1}{2}\left[ {\cos \left( {\alpha + 2\beta } \right) - \cos \alpha } \right]}}\\ = \dfrac{{\sin \left( {\alpha + 2\beta } \right) + \sin \alpha }}{{\cos \left( {\alpha + 2\beta } \right) + \cos \alpha }}\\ = \dfrac{{2\sin \left( {\alpha + \beta } \right)\cos \beta }}{{2\cos \left( {\alpha + \beta } \right)\cos \beta }} = \tan \left( {\alpha + \beta } \right)\end{array}\)