A.To perform
B.Be performed
C.Performed
D.Having performed

A.although
B.in spite of
C.because
D.because of

A.\(\sin A =  - \,\sin \left( {2A + B + C} \right)\)
B.\(\sin A =  - \cos \dfrac{{3A + B + C}}{2}\)
C.\(\cos C = \sin \dfrac{{A + B + 3C}}{2}\)
D.\(\sin \left( {A + B + 2C} \right) =  - \sin C\)

A.\(\sin \dfrac{{A + C}}{2} = \cos \dfrac{B}{2}\)
B.\(\cos \dfrac{{A + C}}{2} = \sin \dfrac{B}{2}\)
C.\(\sin \left( {A + B} \right) = \sin C\)
D.\(\cos \left( {A + B} \right) = \cos C\)

A.\(\sin C =  - \,\sin \left( {A + B} \right)\)
B.\(\cos C = \cos \left( {A + B} \right)\)
C.\(\tan C = \tan \left( {A + B} \right)\)
D.\(\cot C =  - \,\cot \left( {A + B} \right)\)

A.\(\sin \left( {A + C} \right) =  - \sin B\)
B.\(\cos \left( {A + C} \right) =  - \cos B\)
C.\(\tan \left( {A + C} \right) = \tan B\)
D.\(\cot \left( {A + C} \right) = \cot B\)

A.\(\sin \left( {A + B} \right) = \cos C\)
B.\(\cos A = \sin B\)
C.\(\tan A = \cot \left( {B + \dfrac{\pi }{2}} \right)\)
D.\(\cos \dfrac{{A + B}}{2} = \sin \dfrac{C}{2}\)

A.\(\sin \left( {\dfrac{{A + B}}{2}} \right) = \cos \dfrac{C}{2}\)
B.\(\sin \left( {\dfrac{{A + B}}{2}} \right) =  - \cos \dfrac{C}{2}\)
C.\(\sin \left( {\dfrac{{A + B}}{2}} \right) = \sin \dfrac{C}{2}\)
D.\(\sin \left( {\dfrac{{A + B}}{2}} \right) =  - \sin \dfrac{C}{2}\)

A.\(M =  - \cos A\)
B.\(M = \cos A\)
C.\(M = \sin A\)
D.\(M =  - \sin A\)

A.\(\sin \left( {A + B} \right) =  - \sin C\)
B.\(\sin \left( {\dfrac{{A + B}}{2}} \right) = \cos \dfrac{C}{2}\)
C.\(\cos \left( {A + B} \right) = \cos C\)
D.\(\tan \left( {A + B} \right) = \tan C\)