Step 1:
$\tan^{-1}x+\cot^{-1}y=\tan^{-1}3$
$\tan^{-1}x+\tan^{-1}\large\frac{1}{y}$$=\tan^{-1}3$
$\tan^{-1}\bigg[\large\frac{x+1/y}{1-x.\Large\frac{1}{y}}\bigg]$
$\Rightarrow \tan^{-1}3$
$\tan\tan^{-1}\bigg[\large\frac{xy+1}{y-x}\bigg]$$=\tan\tan^{-1}3$
$\Rightarrow \large\frac{xy+1}{y-x}$$=3$
$\Rightarrow xy+1=3(y-x)$
$\Rightarrow xy+1=3y-3x$
$\Rightarrow xy-3y=-3x-1$
$\Rightarrow y[x-3]=-[3x+1]$
$y=\large\frac{-(3x+1)}{x-3}$
$y=\large\frac{3x+1}{3-x}$-----(1)
Step 2:
When $x\rightarrow +ve$ numerator is +ve.
For $y$ to be +ve denominator must be +ve.
(i.e) $3-x>0$
$x< 3\Rightarrow x=1,2$
Substituting the value of $x$ in (1) we get $y=2,7$
$\Rightarrow$ solutions are $x=1,y=2$
$\Rightarrow x=2,y=7$
Hence (a) is the correct answer.