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This is an example question in my book:

To solve for $x$: $$\tan^{-1}\frac{x-1}{x+2}+\tan^{-1}\frac{x+1}{x+2}=\frac{\pi}{4}$$

and it is solved by a direct formula given by $$\tan^{-1}x+\tan^{-1}y=\tan^{-1} \left(\frac{x+y}{1-xy}\right) $$ when $xy<1$. The book uses this formula without checking conditions many many times and it is mildly infuriating. It is alright when checking the condition is simple but in this case finding the range of $\dfrac{x^2-1}{(x+2)^2}$ is a little time consuming and has the square root of 21 in it.

Well my question is:

Why and when can I just peacefully substitute the formula without worrying about the condition as the book leisurely does so?

Blue
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Logan
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    $(x+2)^2$ is positive, so the condition is equivalent to $$x^2-1<(x+2)^2.$$ So the condition is met exactly when $4x>-3.$ So you don’t need to know the range, only the values $x$ that keep your conditions true. – Thomas Andrews Jul 23 '21 at 18:59
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    Short answer: never, unless you have further, perhaps ad-hoc insight into the problem. By ignoring a premise of the formula, the argument is no longer valid, and your conclusion is erroneous. – Theo Bendit Jul 23 '21 at 19:00
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    That said, Cardano's method for solving cubic equations was the result of ignoring assumptions about square roots (that they must be taken of non-negative numbers), and as a result, we ended up with complex numbers. Of course, choosing to ignore assumptions should be done with great care, and some logical framework will be necessary to justify it post-hoc. – Theo Bendit Jul 23 '21 at 19:04
  • @TheoBendit Thank you . I will keep that in mind. – Logan Jul 23 '21 at 19:05
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    You say it is mildly infuriating. So it sounds to me like you are angrily substituting it rather than peacefully. :) – Jair Taylor Jul 23 '21 at 19:08
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    Here, it is still true for any $u,v$ that $$\tan(u+v)=\frac{\tan u+\tan v}{1+\tan u\tan v}$$ So this gives you solutions, and then you need to check them after. This is the same as squaring both sides to the equation, which possibly add roots. You have to check all the solutions you get this way, but it will give you all the solutions. – Thomas Andrews Jul 23 '21 at 19:10
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    If you restrict yourself to real numbers, than perhaps the formula $$\arctan(x)+\arctan(y)=\arctan\left(\frac{x+y}{1-xy}\right)+\pi\operatorname{sign}(x)\mathbf{1}{{xy > 1}},$$ which works for any reals $x, y$ with $xy \neq 1$ might bring peace. Here, $\operatorname{sign}(x)$ denotes the sign of $x$, and $\mathbf{1}{{xy > 1}}$ is the indicator function which takes value $1$ when $xy>1$ and zero otherwise. – Sangchul Lee Jul 23 '21 at 19:13
  • Arctangent addition formula for any range of the arguments – ryang Jul 24 '21 at 03:27

1 Answers1

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We have,

$\mathsf{tan^{-1}\left(\dfrac{x-1}{x+2}\right)+tan^{-1}\left(\dfrac{x+1}{x+2}\right)=\dfrac{\pi}{4}}$

$\mathsf{\implies\,tan^{-1}\left(\dfrac{x+1}{x+2}\right)=\dfrac{\pi}{4}-tan^{-1}\left(\dfrac{x-1}{x+2}\right)}$

$\mathsf{\implies\,tan\left\{tan^{-1}\left(\dfrac{x+1}{x+2}\right)\right\}=tan\left\{\dfrac{\pi}{4}-tan^{-1}\left(\dfrac{x-1}{x+2}\right)\right\}}$

$\mathsf{\implies\,\dfrac{x+1}{x+2}=\dfrac{1-\dfrac{x-1}{x+2}}{1+\dfrac{x-1}{x+2}}}$

$\mathsf{\implies\,\dfrac{x+1}{x+2}=\dfrac{x+2-x+1}{x+2+x-1}}$

$\mathsf{\implies\,\dfrac{x+1}{x+2}=\dfrac{3}{2x+1}}$

$\mathsf{\implies\,(x+1)(2x+1)=3(x+2)}$

$\mathsf{\implies\,2x^2+2x+x+1=3x+6}$

$\mathsf{\implies\,2x^2+3x+1=3x+6}$

$\mathsf{\implies\,2x^2=5}$

$\mathsf{\implies\,x=\pm\sqrt{\dfrac{5}{2}}}$

Now, check the value(s) of x satisfying the given equation,

Those value(s) of x are not satisfying the given equation, will be neglected

There is no need to work with the conditions when solving ITF equations.

In this equation $\mathsf{x=-\sqrt{\dfrac{5}{2}}}$ does not satisfy the given equation

Thus, $\mathsf{x=\sqrt{\dfrac{5}{2}}}$ is the only real solution.

Parthib Ghosh
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