Question

if f(x) = log1-x/1+x so prove that f(x)+f(y) = x+y/1+xy

Answer 1

f(x)=log[1+x/1−x]

f(y)=log[1+y/1−y]

Taking LHS
f(x)+f(y)=log[1+x/1−x]+log[1+y/1−y]
=log(1+x)−log(1−x)+log(1+y)−log(1−y) [log(m/n)=logm−logn]
=log(1+x)+log(1+y)−[log(1−x)+log(1−y)]
=log[(1+x)(1+y)]−log[(1−x)(1−y)]
=log(1+x+y+xy)−log(1−y−x+xy)
=log(1+x+y+xy/1−y−x+xy)

Taking RHS
f(x+y/1+xy)
=log[(1+ x+y/1+xy)/ 1− (x+y/1+xy)]
=log[((1+xy+x+y)/1+xy)/ ((1+xy−x−y)1+xy)]
=log[1+xy+x+y/ 1+xy−x−y]
So, LHS=RHS
Hence proved