Simplifying trigonometric identity with $\arctan$.

Question

How can one prove the following deduction? Assume we know the following result.

$$ \frac{1}{2}\arctan\left( \frac{y}{x+1} \right) + \frac{1}{2}\arctan\left( \frac{y}{x-1} \right) - \arctan\left( \frac{y}{x} \right) = c$$

Then, it is claimed that this is equivalent to the following. I am not able to figure out why.

$$ \frac{(x^2+y^2)^2+y^2-x^2}{xy} = k$$

I am aware of the formula for addition $\arctan(x) + \arctan(y)$ but I am not sure how to deal with prefactors of $1/2$.

Didn't you mean $c$ in the right-hand part of the second equation, not $k$? — Anton Grudkin, Sep 29 '20 at 00:05
@Anton Grudkin hi, no. In general constant might be different (according to the source) — Daniels Krimans, Sep 29 '20 at 01:05

score 2 · Accepted Answer · answered Sep 29 '20 at 00:57

Hint: Start by letting $$A = \dfrac 12 \arctan \dfrac {y}{x+1}$$ and $$B = \dfrac 12 \arctan \dfrac {y}{x-1}$$ and $$C = \arctan \dfrac yx$$

Then we have $\tan 2A = \dfrac {y}{x+1}$, $\tan 2B = \dfrac {y}{x-1}$, and $\tan C = \dfrac yx$, and $$\tan (2A + 2B) = \dfrac {\tan 2A + \tan 2B}{1-\tan 2A \tan 2B}$$

Plug in the values of $\tan 2A$ and $\tan 2B$, add to the value of $\tan C$, and simplify.

score 2 · Answer 2 · answered Sep 29 '20 at 01:46

Hint

$$-2c=\arctan\dfrac yx-\arctan\dfrac y{x+1}+\arctan\dfrac yx-\arctan\dfrac y{x-1}$$

Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

$$-2c=\arctan\dfrac y{x(x+1)+y^2}-\arctan y{x(x-1)+y^2}+m\pi$$

$$\tan(-2c-m\pi)=\tan\left(\arctan\dfrac y{x(x+1)+y^2}-\arctan\dfrac y{x(x-1)+y^2}\right)$$

Simplifying trigonometric identity with $\arctan$.

2 Answers2