Упрощение тригонометрических выражений. Задача 9

\(\dfrac{{\left( {1 + \sin \alpha } \right){\rm{tg}}\left( {\dfrac{\pi }{4}-\dfrac{\alpha }{2}} \right)}}{{\sin \alpha }} = \dfrac{{\left( {1 + \sin \alpha } \right)\dfrac{{{\rm{tg}}\dfrac{\pi }{4}-{\rm{tg}}\dfrac{\alpha }{2}}}{{1 + {\rm{tg}}\dfrac{\pi }{4} \cdot {\rm{tg}}\dfrac{\alpha }{2}}}}}{{\sin \alpha }} = \dfrac{{\left( {1 + \sin \alpha } \right) \cdot \dfrac{{1-{\rm{tg}}\dfrac{\alpha }{2}}}{{1 + {\rm{tg}}\dfrac{\alpha }{2}}}}}{{\sin \alpha }} = \)

\( = \dfrac{{\left( {1 + \sin \alpha } \right) \cdot \dfrac{{\cos \dfrac{\alpha }{2}-\sin \dfrac{\alpha }{2}}}{{\cos \dfrac{\alpha }{2} + \sin \dfrac{\alpha }{2}}}}}{{\sin \alpha }} = \dfrac{{\left( {1 + \sin \alpha } \right)}}{{\sin \alpha }} \cdot \dfrac{{\left( {\cos \dfrac{\alpha }{2}-\sin \dfrac{\alpha }{2}} \right)\left( {\cos \dfrac{\alpha }{2} + \sin \dfrac{\alpha }{2}} \right)}}{{{{\left( {\cos \dfrac{\alpha }{2} + \sin \dfrac{\alpha }{2}} \right)}^2}}} = \)

\( = \dfrac{{1 + \sin \alpha }}{{\sin \alpha }} \cdot \dfrac{{{{\cos }^2}\dfrac{\alpha }{2}-{{\sin }^2}\dfrac{\alpha }{2}}}{{{{\cos }^2}\dfrac{\alpha }{2} + 2\cos \dfrac{\alpha }{2}\sin \dfrac{\alpha }{2} + {{\sin }^2}\dfrac{\alpha }{2}}} = \dfrac{{1 + \sin \alpha }}{{\sin \alpha }} \cdot \dfrac{{\cos \alpha }}{{1 + \sin \alpha }} = \dfrac{{\cos \alpha }}{{\sin \alpha }} = {\rm{ctg}}\alpha .\)