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A question is given in my book.

Solve for $x$:
$$\tan^{-1}(2x)+\tan^{-1}(3x)=\frac\pi4$$

I understood the solution of this question which is as follows:

We have,

$$\begin{align} &\tan^{-1}(2x)+\tan^{-1}(3x) =\frac\pi4 \tag1\\[6pt] \implies\quad &\tan^{−1}\frac{(2x)+(3x)}{1-(2x)(3x)} = 1, \text{if}\; (2x)(3x) <1 \tag2\\[6pt] \implies\quad &\frac{5x}{1-6x^2}=1 \;\text{and}\; x^2<\frac16 \tag3\\[6pt] \implies\quad &(6x-1)(x+1)=0 \;\text{and}\; -\frac{1}{\sqrt{6}}<x<\frac1{\sqrt{6}} \tag4\\[6pt] \implies\quad &x=-1,\frac16 \;\text{and}\; -\frac{1}{\sqrt{6}}<x<\frac1{\sqrt{6}} \tag5\\[6pt] \implies\quad &x=\frac16 \tag6 \end{align}$$

My doubt is that other two Properties of tan; i.e.,

$$\tan^{−1}x + \tan^{−1}y = \begin{cases} \phantom{-}\pi+\tan^{−1}\dfrac{x+y}{1−xy}, &\text{for}\; xy>1, x>0, y>0 \\[6pt] -\pi+\tan^{−1}\dfrac{x+y}{1−xy}, &\text{for}\; xy>1, x<0, y<0 \end{cases}$$

Why we have not used these two properties at step $(2)$?

As it may be possible that some value of $x$ satisfies the condition of these properties and we get one more solution of this equation. Am I thinking wrong so? Please tell me. And if I am right, then in book why only one property has been used?

Blue
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    Please use mathjax, https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference, to properly format the mathematical symbols involved in the question. As it stays, it is hard to read the question. Could you please try to edit the text of the question?! – dan_fulea May 11 '21 at 17:00
  • https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee May 11 '21 at 17:57

1 Answers1

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For any real number $t$, whether $t = \frac{x+y}{1-xy}$ or $t = \frac{2x+3x}{1-(2x)(3x)}$ or something else, $\tan^{-1}(t)$ is always an angle within the interval $\left(-\frac\pi2, \frac\pi2\right)$, that is, $$ -\frac\pi2 < \tan^{-1}(t) < \frac\pi2.$$

It is therefore not possible that $$ \pi + \tan^{-1}(t) = \frac\pi4.$$ That equation could only be satisfied if $\tan^{-1}(t) = -\frac{3\pi}{4},$ but the $-\frac{3\pi}{4}$ is not a possible value of $\tan^{-1}(t)$ for any real $t.$

It is also not possible that $$ -\pi + \tan^{-1}(t) = \frac\pi4.$$

Therefore the only possible case is the one where $$ \tan^{-1}(t) = \frac\pi4.$$ That is, the only possible solution is $t = 1.$

Of course you must then figure out if there is an $x$ that fits your problem such that $t = 1,$ but you seem to have worked that part out already.

But I would approach it this way:

Let $\alpha = \tan^{-1}(2x)$. Let $\beta = \tan^{-1}(3x)$.

Then $\frac\pi4 = \alpha + \beta $ and therefore

$$ \tan\left(\frac\pi4\right) = \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)}. $$

But $\tan(\alpha) = \tan\left(\tan^{-1}(2x)\right) = 2x$, and so forth.

This way we never use the formula for adding two arc tangents, so we never even have to think about its three cases, much less explain why two of them do not apply.

Note that the order of function applications is important here: $\tan\left(\tan^{-1}(t)\right) = t$ always, but $\tan^{-1}\left(\tan(\theta)\right)$ is not always equal to $\theta$. Do you see why?

David K
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  • tan(tan−1(t))=t always, but tan−1(tan(θ)) is not always equal to θ. Can u please tell me why is it so?? – Invisible May 11 '21 at 18:32
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    The definition of $\tan^{-1}(t)$ says it is an angle whose tangent is $t$. More specifically, it says it is the angle between $-\frac\pi2$ and $\frac\pi2$ whose tangent is $t$. But "$\theta$ is an angle whose tangent is $t$" is just a wordy way of saying that $\tan(\theta) = t$. It's just as true if you write $\tan^{-1}(t)$ instead of $\theta$. On the other hand, there are infinitely many angles whose tangent is $t$, and $\tan^{-1}(t)$ is only one of them. For example, $\tan(-3\pi/4) = 1,$ but $\tan^{-1}(1) = \pi/4 \neq -3\pi/4.$ – David K May 12 '21 at 02:25