<s>[INST] trigonometric equation for tan^(-1) Need to solve this equation:
$$\tan^{-1}\frac{x}{3} +\tan^{-1}x=  \tan^{-1}2.$$
Method with explanation is highly appreciated 
 [/INST] 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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