Đề bài

Tính \(\dfrac{{\sin \alpha  + \sin \beta c{\rm{os}}\left( {\alpha  + \beta } \right)}}{{\cos \alpha  - \sin \beta \sin \left( {\alpha  + \beta } \right)}}\)

• A. $\tan \left( {\alpha  + \beta } \right)$
• B. $\cot \left( {\alpha  + \beta } \right)$
• C. $\sin \left( {\alpha  + \beta } \right)$
• D. $\cos \left( {\alpha  + \beta } \right)$

Lời giải của GV Loigiaihay.com

Ta có

\(\dfrac{{\sin \alpha  + \sin \beta \cos \left( {\alpha  + \beta } \right)}}{{\cos \alpha  - \sin \beta \sin \left( {\alpha  + \beta } \right)}}\)\( = \dfrac{{\sin \alpha  + \dfrac{1}{2}\left[ {\sin \left( {\alpha  + 2\beta } \right) + \sin \left( { - \alpha } \right)} \right]}}{{\cos \alpha  + \dfrac{1}{2}\left[ {\cos \left( {\alpha  + 2\beta } \right) - \cos \left( { - \alpha } \right)} \right]}}\) \( = \dfrac{{\sin \alpha  + \dfrac{1}{2}\left[ {\sin \left( {\alpha  + 2\beta } \right) - \sin \alpha } \right]}}{{\cos \alpha  + \dfrac{1}{2}\left[ {\cos \left( {\alpha  + 2\beta } \right) - \cos \alpha } \right]}}\)\( = \dfrac{{\sin \alpha  + \dfrac{1}{2}\sin \left( {\alpha  + 2\beta } \right) - \dfrac{1}{2}\sin \alpha }}{{\cos \alpha  + \dfrac{1}{2}\cos \left( {\alpha  + 2\beta } \right) - \dfrac{1}{2}\cos \alpha }}\)\( = \dfrac{{\dfrac{1}{2}\sin \alpha  + \dfrac{1}{2}\sin \left( {\alpha  + 2\beta } \right)}}{{\dfrac{1}{2}\cos \alpha  + \dfrac{1}{2}\cos \left( {\alpha  + 2\beta } \right)}}\) \( = \dfrac{{\sin \left( {\alpha  + 2\beta } \right) + \sin \alpha }}{{\cos \left( {\alpha  + 2\beta } \right) + \cos \alpha }}\) \( = \dfrac{{2\sin \dfrac{{\left( {\alpha  + 2\beta  + \alpha } \right)}}{2}\cos \dfrac{{\left( {\alpha  + 2\beta  - \alpha } \right)}}{2}}}{{2\cos \dfrac{{\left( {\alpha  + 2\beta  + \alpha } \right)}}{2}\cos \dfrac{{\left( {\alpha  + 2\beta  - \alpha } \right)}}{2}}}\)\( = \dfrac{{2\sin \left( {\alpha  + \beta } \right)\cos \beta }}{{2\cos \left( {\alpha  + \beta } \right)\cos \beta }} = \tan \left( {\alpha  + \beta } \right)\)

Đáp án : A