If $$f(x) = \log \frac{(1+x)}{(1-x)}$$, then f(x) + f(y) is

If $$f(x) = \log \frac{(1+x)}{(1-x)}$$ then $$f(y) = \log \frac{(1+y)}{(1-y)}$$

Also Log (A*B)= Log A + Log B 

f(x)+f(y) = $$ \log \frac{(1+x)(1+y)}{(1-x)(1-y)}$$ 

=$$\log\frac{\left(1+xy\ +x\ +y\right)}{\left(1+xy-x-y\right)}$$ 

Dividing numberator and denominator by (1+xy)

$$\log\frac{\frac{\left(1+xy\ +x\ +y\right)}{1+xy}}{\frac{\left(1+xy-x-y\right)}{1+xy}}$$

 =$$\log\frac{\frac{1+xy\ }{1+xy}+\frac{\left(x+y\right)}{1+xy}}{\frac{1+xy\ }{1+xy}-\frac{\left(x+y\right)}{1+xy}}$$

= $$\log { \frac{1+ \frac{(x+y)}{(1+xy)}}{1- \frac{(x+y)}{(1+xy)}}}$$

Hence option B.