[tex]Se \;da: \\ tg\,a= \sqrt{3} \\ \\ Se \;cere \;sa \;se\;arate\;ca: \\ \frac{sin\,a-cos\,a}{cos\,a+sin\,a}=2- \sqrt{3} \\ \\ Rezolvare: \\ \frac{sin\,a-cos\,a}{cos\,a+sin\,a}= \frac{sin\,a-cos\,a}{sin\,a+cos\,a}=\;\; \text{simplificam fortat cu }\;\; cos\,a\;\;\;= [/tex]
[tex]= \frac{\frac{sin\,a}{cos\,a}- \frac{cos\,a}{cos\,a}}{\frac{sin\,a}{cos\,a}+ \frac{cos\,a}{cos\,a}} = \frac{tg\,a -1}{tg\,a +1} = \\ \\ = \frac{ \sqrt{3} -1}{\sqrt{3} +1}= rationalizam = \frac{ (\sqrt{3} -1)(\sqrt{3} -1)}{(\sqrt{3} +1)(\sqrt{3} -1)}= \\ \\ = \frac{ (3 -2\sqrt{3} +1)}{3-1}=\frac{ (4 -2\sqrt{3} )}{2}=\frac{ 2(2 -\sqrt{3} )}{2}= \boxed{2 -\sqrt{3}} \\ \\ c.c.t.d.[/tex]
