Question

Prove that f(a)+f(b)=f(a+b/1+ab) when f(x)=log(1-x/1+x)

Answer 1

f(x)=log[1+x/1−x]

so, f(y)=log[1+y/1−y]

Consider, 
LHS=f(x)+f(y)=log[1+x1−x]+log[1+y1−y]=log(1+x)−log(1−x)+log(1+y)−log(1−y)     

[log(mn)=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+xy1−y−x+xy)

RHS

= f(x+y1+xy)

=log⎡⎣1+x+y1+xy1−x+y1+xy⎤⎦

=log⎡⎣1+xy+x+y1+xy1+xy−x−y1+xy⎤⎦

=log[1+xy+x+y1+xy−x−y]

So, LHS=RHSHence proved