zakinutuzi

2022-01-03

How to show that

$\mathrm{log}(\frac{2a}{1-{a}^{2}}+\frac{2b}{1-{b}^{2}}+\frac{2c}{1-{c}^{2}})=\mathrm{log}\frac{2a}{1-{a}^{2}}+\mathrm{log}\frac{2b}{1-{b}^{2}}+\mathrm{log}\frac{2c}{1-{c}^{2}}$

Marcus Herman

Beginner2022-01-04Added 41 answers

HINT:

Use$\mathrm{log}a+\mathrm{log}b+\mathrm{log}c=\mathrm{log}\left(abc\right)$

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

$\sum \mathrm{tan}A=\prod \mathrm{tan}A\Rightarrow A+B+C=n\pi$ where $n$ is any integer

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

$\Rightarrow \sum \mathrm{tan}2A=\prod \mathrm{tan}2A\text{}\text{as}\text{}2A+2B+2C=2n\pi$

Now apply logarithm now

Use

and then put

and

Now apply logarithm now

reinosodairyshm

Beginner2022-01-05Added 36 answers

OK, heres

Vasquez

Expert2022-01-09Added 669 answers

The expression

Here I'm using capital A to refer to the angle and lower-case a to refer to its tangent:

The identity you're trying to prove is equivalent to

(with lower-case a,b,c), so that's the same as

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

and in the second place, if

So what you're being asked to prove is that if

We've see that the fact that

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

$\frac{20b}{{\left(4{b}^{3}\right)}^{3}}$

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