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Prove that

$$\tan^{-1}x + \tan^{-1}\frac{2x}{1-x^2}=\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)\;,\;\; |x| < \frac{1}{\sqrt{3}}$$

By taking R.H.S $\tan^{-1}(\frac{3x-x^3}{1-3x^2}) = \tan^{-1}(\frac{3\tan\theta -\tan^{3}\theta}{1-3\tan^2\theta})$ where $\tan\theta =x$

$3\tan^{-1}x = \tan^{-1}x + \tan^{-1}\frac{2x}{1-x^2}$ = L.H.S.

My question is what is the role of $|x| < \frac{1}{\sqrt{3}}$ in this question.... please guide...

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4 Answers 4

$|x|<1/\sqrt3$ gives you necessary condition so that you substitute $tan\theta=x$ and still compute RHS as $3\theta$ otherwise there would have been a angle shift of $\pi$ towards left.

It's because the range of $tan^{-1}\in[-\pi/2,\pi/2]$ but the LHS of the equation has range$\in [-\pi,\pi]$ and thus the RHS will compute to LHS-$\pi$ $\forall x\ge 1/\sqrt3$

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An idea: differentiate both sides (I use $\arctan\,$ instead of $\,\tan^{-1}\,$:

$$\text{I}\;\;\;\;\;\;\;\left(\arctan x+\arctan\frac{2x}{1-x^2}\right)'=\frac{1}{1+x^2}+\frac{2(1+x^2)}{\color{green}{(1-x^2)^2}}\frac{1}{1+\frac{4x^2}{\color{green}{(1-x^2)^2}}}=$$

$$=\frac{1}{1+x^2}+\frac{2(1+x^2)}{(1+x^2)^2}=\frac{3}{1+x^2}$$

$$\text{II}\;\;\;\;\left(\arctan\frac{3x-x^3}{1-3x^2}\right)'=\frac{3(x^2-1)(3x^2+1)}{\color{red}{(3x^2-1)^2}}\frac{1}{1+\frac{(3x-x^3)^2}{\color{red}{(3x^2-1)^2}}}=$$

$$=\frac{3(x^2-1)(3x^2+1)}{(x+1)^3}=\frac{3(x-1)(3x^2+1)}{(x+1)^2}$$

so unless I made a mistake (nothing rare...), both functions cannot be equal in the given interval (or anywhere else) since their derivatives aren't...

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In simple terms, your RHS is not "well defined" for $|x|=1/\sqrt{3}$. For $|x| > 1/\sqrt{3}$, your RHS could be negative while your LHS is positive. This is taking the domain and range of $\tan(x)$ into consideration.

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First of all, all the inverse trigonometric functions are multivalued.

But the given inequality implicitly assumes the principal values of $\arctan$

Let $\arctan x=\theta$ where $-\frac\pi2\le \theta\le \frac\pi2$

So, $\tan\theta=x$

Now, $\frac{2x}{1-x^2}=\tan2\theta$ and $-\pi\le2\theta\le \pi$

If $-\pi\le 2\theta<-\frac\pi2\iff -\frac\pi2\le\theta<-\frac\pi4, \arctan \left(\frac{2x}{1-x^2}\right)=2\theta+\pi=2\arctan x+\pi$ to fit the value in $[-\frac\pi2,\frac\pi2]$

If $-\frac\pi2\le 2\theta\le\frac\pi2\iff -\frac\pi4\le\theta\le\frac\pi4, \arctan \left(\frac{2x}{1-x^2}\right)=2\theta=2\arctan x$

If $\frac\pi2<2\theta\le\pi\iff \frac\pi4<\theta\le\frac\pi2, \arctan \left(\frac{2x}{1-x^2}\right)=2\theta-\pi=2\arctan x-\pi$

Again, $\frac{3x-x^3}{1-3x^2}=\frac{3\theta-\tan^3\theta}{1-3\tan^2\theta}=\tan3\theta$ and $-\frac{3\pi}2\le 3\theta\le \frac{3\pi}2$

Applying the previous method, $$\arctan \left(\frac{3x-x^3}{1-3x^2}\right)= \begin{cases} 3\theta+\pi &\mbox{if } -\frac{3\pi}2\le3\theta<-\frac\pi2\iff -\frac\pi2\le\theta<-\frac\pi6\\ 3\theta &\mbox{if } -\frac\pi2\le3\theta\le \frac\pi2\iff -\frac\pi6\le\theta\le \frac\pi6\\ 3\theta-\pi &\mbox{if } \frac\pi2<3\theta\le \frac{3\pi}2\iff \frac\pi6<\theta\le\frac\pi2 \end{cases} $$

So, we need to test for the following ranges:

$$\left[-\frac\pi2,-\frac\pi4\right); \left[-\frac\pi4,-\frac\pi6\right);\left[-\frac\pi6,\frac\pi6\right]; \left(\frac\pi6,\frac\pi4\right];\left(\frac\pi4,\frac\pi2\right]$$

Observe that the given inequality holds if $\theta$ lies $\in \left[-\frac\pi6,\frac\pi6\right]$

$\implies \tan\left(-\frac\pi6 \right)\le \tan\theta\le \tan\left(\frac\pi6 \right)$

$\implies -\frac1{\sqrt3}\le x\le \frac1{\sqrt3}$

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