Réponses

2014-06-19T23:06:42+02:00
Bonsoir,

\dfrac{\sin a-\sin b}{\sin a+\sin b}=\dfrac{2\sin\dfrac{a-b}{2}\cos\dfrac{a+b}{2}}{2\sin\dfrac{a+b}{2}\cos\dfrac{a-b}{2}}\\\\\\=\dfrac{\sin\dfrac{a-b}{2}\cos\dfrac{a+b}{2}}{\sin\dfrac{a+b}{2}\cos\dfrac{a-b}{2}}\\\\\\=\dfrac{\sin\dfrac{a-b}{2}}{\cos\dfrac{a-b}{2}}\times\dfrac{\cos\dfrac{a+b}{2}}{\sin\dfrac{a+b}{2}}\\\\

=\tan \dfrac{a-b}{2}\times\dfrac{1}{\dfrac{\sin\dfrac{a+b}{2}}{\cos\dfrac{a+b}{2}}}\\\\\\=\tan \dfrac{a-b}{2}\times\dfrac{1}{\tan \dfrac{a+b}{2}}\\\\\\=\boxed{\dfrac{\tan \dfrac{a-b}{2}}{\tan \dfrac{a+b}{2}}}
vache de simplification ! ;-)
Merci beaucoup :)
Avec plaisir :)