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If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$$

Trial: If put $a=\frac{2a}{1-a^2}$ and the similar we are done. Can we do this in this way? Otherwise how to do. Please help.

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Are you sure it's correct? $\log(1+1+1)\neq 0$ but $\log(1)+\log(1)+\log(1)= 0$ –  freak_warrior Nov 16 '13 at 16:57
    
@freak_warrior : I would take the question to mean this: For some values of $a,b,c$, the equality holds, and the equality to be shown is something that is true in those cases. It's as if we'd said "If $ab=a+b$, then...". It is not true in general that $ab=a+b$, but there are some particular cases in which it holds, for example $a=3$ and $b=1.5$. –  Michael Hardy Nov 16 '13 at 17:21
    
When you see the expression $\dfrac{2a}{1-a^2}$, you should think of the tangent function. –  Michael Hardy Nov 16 '13 at 17:22
    
I've now posted two answers to this question, one using trigonometry and the other not using trigonometry. –  Michael Hardy Nov 16 '13 at 23:32

3 Answers 3

HINT:

Use $\log a+\log b+\log c=\log(abc) $

and then put $a=\tan A$ etc. to find

$\displaystyle \sum\tan A=\prod\tan A\implies A+B+C=n\pi$ where $n$ is any integer (See here)

and $\displaystyle\frac{2a}{1-a^2}=\frac{2\tan A}{1-\tan^2A}=\tan2A$

$\displaystyle\implies \sum\tan2A=\prod\tan2A$ as $2A+2B+2C=2n\pi$

Now apply logarithm now

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OK, here's another way to do it:

Let $a\circ b=\dfrac{a+b}{1-ab}$.

Prove that this operation is associative.

Show that $a\circ b\circ c = \dfrac{a+b+c-abc}{1-ab-ac-bc}$.

That implies that if $a+b+c=abc$, then $a\circ b\circ c=0$.

It's easy to show that $0\circ 0=0$, so we have

\begin{align} 0 & = 0\circ 0 = (a\circ b\circ c)\circ(a\circ b\circ c) \\[12pt] & = (a\circ a)\circ(b\circ b)\circ(c\circ c) \qquad (\text{This is where associativity is used.}) \\[12pt] & = \frac{2a}{1-a^2}\circ\frac{2b}{1-b^2} \circ\frac{2c}{1-c^2}. \end{align}

So that implies the numerator $\dfrac{2a}{1-a^2}+\dfrac{2b}{1-b^2} +\dfrac{2c}{1-c^2} -\dfrac{2a}{1-a^2}\cdot\dfrac{2b}{1-b^2} \cdot\dfrac{2c}{1-c^2}$, is $0$.

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The expression $\dfrac{2a}{1-a^2}$ should make you think of the tangent function, as in $$\tan(2A)=\frac{2\tan A}{1-(\tan A)^2}$$

Here I'm using capital $A$ to refer to the angle and lower-case $a$ to refer to its tangent: $a=\tan A$.

$\log(a+b+c)=\log a+\log b+\log c$ implies $\log(a+b+c)=\log(abc)$, which then implies $a+b+c=abc$, and following the pattern in capital and lower-case letters above, we have $\tan A+\tan B+\tan C=\tan A\tan B\tan C$.

The identity you're trying to prove is equivalent to $$ \frac{2a}{1-a^2} + \frac{2b}{1-b^2}+\frac{2c}{1-c^2} = \frac{2a}{1-a^2} \cdot \frac{2b}{1-b^2}\cdot\frac{2c}{1-c^2} $$ (with lower-case $a,b,c$), so that's the same as $$ \tan(2A)+\tan(2B)+\tan(2C)=\tan(2A)\tan(2B)\tan(2C). $$

Now at this point I might not know how to proceed further if I hadn't seen the following at some point in the past. In the first place, the usual formula for the tangent of a sum implies with 30 seconds' more work that $$ \tan(A+B+C)=\frac{\tan A+\tan B+\tan C-\tan A\tan B\tan C}{1-\tan A\tan B-\tan A\tan C-\tan B\tan C}, $$ and in the second place, if $A+B+C=\pi\text{ radians or }180^\circ\text{ or a half-circle}$ then $\tan(A+B+C)=0$. So the fraction is $0$ and therefore the numerator is $0$ and therefore $\tan A+\tan B+\tan C=\tan A\tan B\tan C$.

So what you're being asked to prove is that if $\tan A+\tan B+\tan C=\tan A\tan B\tan C$ then $\tan(2A)+\tan(2B)+\tan(2C)=\tan(2A)\tan(2B)\tan(2C)$.

We've see that the fact that $A+B+C=\text{a half circle}$ implies the first of these identities only because it implies that $\tan(A+B+C)=0$. So you really just need to show that if $\tan(A+B+C)=0$ then $\tan(2A+2B+2C)=0$. It's not hard to show that that's the same as saying that if $A+B+C$ is an integer multiple of a half-circle, then so is $2A+2B+2C$.

Moral: Don't do this problem if you've forgotten your trigonometry.

However, since the problem as stated doesn't mention trigonometry, I'm wondering if there's another way to do it that avoids that. Probably there is. The tangent function and the identities we used involve a particular way of parametrizing the circle, as $A\mapsto(\cos A,\sin A)$. But the result doesn't seem to be one that should depend on such a choice of parametrization.

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