If \(\frac{{x + 1}}{{x - 1}}{ \text{ = }}\frac{a}{b} \) and \(\frac{{1 - y}}{{1 + y}}{ \text{ = }}\frac{b}{a}{ \text{,}} \) then the value of \(\frac{{x - y}}{{1 + xy}}\) = ?
-
1]
\(\frac{{2ab}}{{{a^2} - {b^2}}}\)
-
2]
\(\frac{{{a^2} - {b^2}}}{{2ab}}\)
-
3]
\(\frac{{{a^2} + {b^2}}}{{2ab}}\)
-
4]
\(\frac{{{a^2} - {b^2}}}{{ab}}\)
Solution
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where x,y,z are natural member , then find x+y+z are:-
=3.605 and 
=11.40 , Find the value of 
+ 
+