Question

tg(a)=1/3 find (sina-cosa)/(sin(a)+cos(a))

Answer 1

\[\tan a = \frac{\sin a }{\cos a}=\frac{1}{3}\]\[\frac{\sin a-\cos a}{\sin a +\cos a}=\frac{1}{1+\frac{1}{\tan a}}-\frac{1}{\tan a +1}=\frac{1}{1+3}-\frac{1}{\frac{1}{3} +1}=-\frac{1}{2}\]

Answer 2

im sry but i dont get it

Answer 3

what did you exactly substituted?

Answer 4

OK. I've added a few more details inbetween:
\[\frac{\sin a-\cos a}{\sin a +\cos a}=\frac{\sin a}{\sin a + \cos a}-\frac{\cos a}{\sin a + \cos a}=\]\[=\frac{\sin a}{\sin a + \cos a}\cdot \frac{\frac{1}{\sin a}}{\frac{1}{\sin a}}-\frac{\cos a}{\sin a + \cos a}\cdot \frac{\frac{1}{\cos a}}{\frac{1}{\cos a}}=\]
\[=\frac{1}{1+\frac{\cos a }{\sin a}}-\frac{1}{\frac{\sin a}{\cos a}+1}=\]
\[=\frac{1}{1+\frac{1}{\tan a}}-\frac{1}{\tan a +1}=\frac{1}{1+3}-\frac{1}{\frac{1}{3} +1}=-\frac{1}{2}\]