Expansion of $f (x,y) = tan^{-1} \frac{⁡y}{x}$ upto first degree containing (x+1) & (y-1) is __________

Question

Expansion of  $f (x,y) = tan^{-1} \frac{⁡y}{x}$ upto first degree containing (x+1) & (y-1) is __________

TuteeHUB · Accepted Answer

NA

TuteeHUB · Answer

$\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}  \frac{-1}{2}  + \frac{(y-1)^2}{2!}  \frac{1}{2}$

TuteeHUB · Answer

$\frac{π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}  \frac{1}{4} +  \frac{(y-1)^2}{2!} \frac{1}{4}$

TuteeHUB · Answer

$\frac{5π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}   \frac{-1}{4} + \frac{(y-1)^2}{2!}  \frac{1}{4}$

TuteeHUB · Answer

$\frac{3π}{4}  + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!}   \frac{-1}{4}  + \frac{(y-1)^2}{2!} \frac{1}{4}$

1.	Expansion of \(f (x,y) = tan^{-1} \frac{⁡y}{x}\) upto first degree containing (x+1) & (y-1) is __________
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.	\(\frac{3π}{4} + \frac{(x+1)}{1!} \frac{-1}{2} + \frac{(y-1)}{1!} \frac{-1}{2} + \frac{(x+1)^2}{2!} \frac{-1}{4} + \frac{(y-1)^2}{2!} \frac{1}{4}\)
Answer» B.

Expansion of \(f (x,y) = tan^{-1} \frac{⁡y}{x}\) upto first degree containing (x+1) & (y-1) is __________