 How to show that \log(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})=\log\frac{2a}{1-a^2}+\log\frac{2b}{1-b^2}+\log\frac{2c}{1-c^2} zakinutuzi 2022-01-03 Answered
How to show that
$$\displaystyle{\log{{\left({\frac{{{2}{a}}}{{{1}-{a}^{{2}}}}}+{\frac{{{2}{b}}}{{{1}-{b}^{{2}}}}}+{\frac{{{2}{c}}}{{{1}-{c}^{{2}}}}}\right)}}}={\log{{\frac{{{2}{a}}}{{{1}-{a}^{{2}}}}}}}+{\log{{\frac{{{2}{b}}}{{{1}-{b}^{{2}}}}}}}+{\log{{\frac{{{2}{c}}}{{{1}-{c}^{{2}}}}}}}$$

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HINT:
Use $$\displaystyle{\log{{a}}}+{\log{{b}}}+{\log{{c}}}={\log{{\left({a}{b}{c}\right)}}}$$
and then put $$\displaystyle{a}={\tan{{A}}}$$ etc. to find
$$\displaystyle\sum{\tan{{A}}}=\prod{\tan{{A}}}\Rightarrow{A}+{B}+{C}={n}\pi$$ where $$\displaystyle{n}$$ is any integer
and $$\displaystyle{\frac{{{2}{a}}}{{{1}−{a}^{{2}}}}}={\frac{{{2}{\tan{{A}}}}}{{{1}−{{\tan}^{{2}}{A}}}}}={\tan{{2}}}{A}$$
$$\displaystyle\Rightarrow\sum{\tan{{2}}}{A}=\prod{\tan{{2}}}{A}\ \text{ as }\ {2}{A}+{2}{B}+{2}{C}={2}{n}\pi$$
Now apply logarithm now
Not exactly what you’re looking for? reinosodairyshm
OK, here's another way to do it:
Let $$\displaystyle{a}\circ{b}={\frac{{{a}+{b}}}{{{1}−{a}{b}}}}.$$
Prove that this operation is associative.
Show that $$\displaystyle{a}\circ{b}\circ{c}={\frac{{{a}+{b}+{c}−{a}{b}{c}}}{{{1}−{a}{b}−{a}{c}−{b}{c}}}}.$$
That implies that if $$\displaystyle{a}+{b}+{c}={a}{b}{c}$$, then $$\displaystyle{a}\circ{b}\circ{c}={0}.$$
It's easy to show that $$\displaystyle{0}\circ{0}={0}$$, so we have
$$\displaystyle{0}={0}\circ{0}={\left({a}\circ{b}\circ{c}\right)}\circ{\left({a}\circ{b}\circ{c}\right)}$$
$$\displaystyle={\left({a}\circ{a}\right)}\circ{\left({b}\circ{b}\right)}\circ{\left({c}\circ{c}\right)}$$ (This is where associativity is used.)
$$\displaystyle={\frac{{{2}{a}}}{{{1}−{a}^{{2}}}}}\circ{\frac{{{2}{b}}}{{{1}−{b}^{{2}}}}}\circ{\frac{{{2}{c}}}{{{1}−{c}^{{2}}}}}.$$
So that implies the numerator
$$\displaystyle{\frac{{{2}{a}}}{{{1}−{a}^{{2}}}}}+{\frac{{{2}{b}}}{{{1}−{b}^{{2}}}}}+{\frac{{{2}{c}}}{{{1}−{c}^{{2}}}}}-{\frac{{{2}{a}}}{{{1}−{a}^{{2}}}}}\cdot{\frac{{{2}{b}}}{{{1}−{b}^{{2}}}}}\cdot{\frac{{{2}{c}}}{{{1}−{c}^{{2}}}}}\text{, is }\ {0}$$ Vasquez

The expression $$\frac{2a}{1-a^2}$$ 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 $$A+B+C=\pi$$ radians or $$180^\circ$$ 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 $$A+B+C=a$$ 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 $$A\rightarrow(\cos A,\sin A)$$. But the result doesn't seem to be one that should depend on such a choice of parametrization.