Expansion of \(f (x,y) = tan^{-1} \frac{y}{x}\) upto first degree containing (x+1) & (y-1) is __________
Posted: Thu Jul 14, 2022 9:29 am
a) \(\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{2} + \frac{(y-1)^2}{2!} \frac{1}{2}\)
b) \(\frac{π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}\)
c) \(\frac{5π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}\)
d) \(\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}\)
b) \(\frac{π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}\)
c) \(\frac{5π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}\)
d) \(\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}\)