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Need to solve this equation: $$\tan^{-1}\frac{x}{3} +\tan^{-1}x= \tan^{-1}2.$$ Method with explanation is highly appreciated

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Apply tangent to both sides and use sum of angles identity for this function:

$$\arctan\frac x3+\arctan x=\arctan2\implies \tan\left(\arctan\frac x3+\arctan x\right)=\tan\arctan 2=2\implies$$

$$\implies\frac{\frac x3+x}{1-\frac{x^2}3}=2\;\;\ldots\;\text{take it from here}\;\ldots$$

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what if $x=-3$? – lab bhattacharjee Feb 15 '14 at 12:19
What with it? I bet the OP can come up with his own conclusion as we must limit the tangent function in order to get its inverse...Or you can give it all to him as I can see you already did. – DonAntonio Feb 15 '14 at 12:44
Do you mean, OP will validate each value against the given relation? – lab bhattacharjee Feb 15 '14 at 12:49
I don't know, @labbhattacharjee. Why don't you ask him? – DonAntonio Feb 15 '14 at 12:50

We need to be careful while summing the inverse trigonometric ratios.

Following this, we gave $$\tan^{-1}\frac x3 +\tan^{-1}x=\begin{cases} \arctan\frac{4x}{3-x^2} &\mbox{if } x\cdot\frac x3<1\iff x^2\le3\iff -\sqrt3\le x\le\sqrt3\\ \pi+\arctan\frac{4x}{3-x^2} & \mbox{ otherwise } \end{cases} $$

Now using this, $\displaystyle -\frac\pi2\le \tan^{-1}2\le\frac\pi2$

More precisely, $\displaystyle 0<\tan^{-1}2<\frac\pi2$

If $\displaystyle-\sqrt3\le x\le\sqrt3, \frac{4x}{3-x^2}=2\iff x^2+2x-3=0\iff x=-3,x=1$

Clearly,$\displaystyle x\ne-3$

which is more evident from $\displaystyle\tan^{-1}(-3)+\tan^{-1}(-1)<0,$ hence $\ne\tan^{-1}2$

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