If $f(x)=\log ((1-x)/(1+x))$ $(-1<x<1)$, show that

Question

If $f(x)=\log \left(\dfrac {1-x}{1+x} \right)$ $(-1<x<1)$, show that $$f \left(\dfrac {2ab}{1+a^2b^2} \right)=2f(ab)$$ Where $|ab|<1$

My Attempt. I got $f \left(\dfrac {2ab}{1+a^2b^2} \right)=-2\log (1) -2\log (1)$

Answer 1

$$f\left(\frac{2ab}{1+a^2b^2}\right)=\ln\frac{1-\frac{2ab}{1+a^2b^2}}{1+\frac{2ab}{1+a^2b^2}}=\ln\left(\frac{1-ab}{1+ab}\right)^2=2\ln\frac{1-ab}{1+ab}=2f(ab)$$ because $\frac{1-ab}{1+ab}>0$.

Answer 2

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

Can you continue?

Answer 3

$$f\left(\dfrac{2t}{1+t^2}\right)=\cdots=\left(\dfrac{1-t}{1+t}\right)^2=2f(t)$$ for $-1<t<1$

Here $t=ab$

Answer 4

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