If $a^2+b^2=1,$ then $\frac{1+b+ia}{1+b-ia}=$

Given that $a^2+b^2=1$, therefore

$\frac{1+b+ia}{1+b-ia}=\frac{(1+b+ia)(1+b+ia)}{(1+b-ia)(1+b+ia)}$

$=\frac{{(1+b)}^2-a^2+2ia(1+b)}{1+b^2+2b+a^2}$$=\frac{(1-a^2)+2b+b^2+2ia(1+b)}{2(1+b)}$

$=\frac{2b^2+2b+2ia(1+b)}{2(1+b)}=b+ia$

Trick :  Put  $a=0,b=1$, $\frac{1+b+ia}{1+b-ia}=\frac{1+1+0}{1+1-0}=1$

But options (1) and (3) give 1.

So again put $a=1,b=0,\frac{1+b+ia}{1+b-ia}=\frac{1+i}{1-i}=i$

Which gives (3) only.