Question

Știind ca tg a=radical din 3 , aratati ca sina-cosa totul supra cosa+sina=2-radical din 3

Answer 1

   
[tex]\displaystyle \\ tg~a= \sqrt{3} \\ \\ \frac{sin~a-cos~a}{cos~a+sin~a} =\frac{sin~a-cos~a}{sin~a+cos~a} = \\ \\ \texttt{(Simplificam fortat cu "cos a". )} \\ \\ =\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} = ~~~~~ \texttt{(Rationalizam numitorul.)}\\ \\ = \frac{( \sqrt{3} -1)( \sqrt{3} -1)}{(\sqrt{3}+1)( \sqrt{3} -1)} = \frac{ (\sqrt{3})^2 -2\sqrt{3} +1}{(\sqrt{3})^2-1}= [/tex]


[tex]\displaystyle \\ =\frac{ 3 -2\sqrt{3} +1}{3-1}= \\ \\ =\frac{ 4 -2\sqrt{3} }{2}= \frac{ 2(2 -\sqrt{3}) }{2}= \boxed{2 -\sqrt{3} }\\ \\ \texttt{cctd} [/tex]