If f(x) = log((1+x)/(1-x)), then f(x)+f(y)=?

(1) f(x+y)
(2) f(1+xy) (3)(x+y) f(1+xy) (4)f((x+y)/(1+xy))

Solution follows here:
Solution:

f(x) = log((1+x)/(1-x))

f(y) = log((1+y)/(1-y))

f(x)+f(y) = log((1+x)/(1-x)) + log((1+y)/(1-y))

Formula here is ---- log(m) + log(n) = log(m*n)

f(x)+f(y) = log ((1+x)/(1-x) * (1+y)/(1-y))

=> f(x)+f(y) = log ((1+xy+x+y)/(1+xy-x-y))

Dividing numerator and denominator (with in log)
by 1+xy:

=> f(x)+f(y) = log ({(1+xy+x+y)/(1+xy)}/{(1+xy-x-y)
/(1+xy)})

=> f(x)+f(y) = log ({1+{(x+y)/(1+xy)}/ {1-{(x+y)/(1+xy)})

=> f(x)+f(y) = f((x+y)/(1+xy))

**Answer(4)**
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