Question

Ex 23 va rog............

Answer 1

 

$\displaystyle\bf\\\frac{cos(x+y)-sin(x-y)}{cos(x-y)-sin(x+y)}=\frac{cos~y+sin~y}{cos~y-siny~}=\frac{1+tg~y}{1-tgy~}\\\\\\\frac{cos(x+y)-sin(x-y)}{cos(x-y)-sin(x+y)}=\\\\\\=\frac{cosxcosy-sinxsiny-(sinxcosy-sinycosx)}{cosxcosy+sinxsiny-(sinxcosy+sinycosx)}=\\\\\\=\frac{cosxcosy-sinxsiny-sinxcosy+sinycosx}{cosxcosy+sinxsiny-sinxcosy-sinycosx}=\\\\\\=\frac{cosx(cosy+siny)-sinx(cosy+siny)}{cosx(cosy-siny)-sinx(cosy-siny)}=$

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$\displaystyle\bf\\=\frac{(cosy+siny)(cosx-sinx)}{(cosy-siny)(cosx-sinx)}=\\\\Simplificam~cu~(cosx-sinx)\\\\=\boxed{\bf\frac{cosy+siny}{cosy-siny}}~~Am~demonstrat~prima~egalitate.\\\\Ne~ocupam~de~egalitatea~a~doua.\\\\\\\frac{cosy+siny}{cosy-siny}=~~Simplificam~fortat~cu~cosy\\\\=\frac{~~\dfrac{cosy+siny}{cosy}~~}{\dfrac{cosy-siny}{cosy}} =\frac{~~\dfrac{cosy}{cosy}+\dfrac{siny}{cosy}~~}{\dfrac{cosy}{cosy}-\dfrac{siny}{cosy}}=\boxed{\bf\frac{1+tg~y}{1-tg~y}}\\\\\\Am~demonstrat~si~a~doua~egalitate.$