trigonometric equation for tan^(-1)

Question

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

Answer 1

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

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

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

Answer 2

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 $\ne\tan^{-1}2$