If $\tan ^{-1}(x+2)+\tan ^{-1}(x-2)-\tan ^{-1}\left(\frac{1}{2}\right)=0$, then one value of $x$ is

If $\tan ^{-1}(x+2)+\tan ^{-1}(x-2)-\tan ^{-1}\left(\frac{1}{2}\right)=0$, then one value of $x$ is
  1. -1
  2. $\frac{1}{2}$
  3. 1
  4. 2

Solution

$\begin{aligned} & \tan ^{-1}(x+2)+\tan ^{-1}(x-2)-\tan ^{-1}\left(\frac{1}{2}\right)=0 \\ & \Rightarrow \tan ^{-1}\left(\frac{(x+2)+(x-2)}{1-(x+2)(x-2)}\right)-\tan ^{-1}\left(\frac{1}{2}\right)=0 \\ & \Rightarrow \tan ^{-1}\left(\frac{2 x}{5-x^2}\right)-\tan ^{-1}\left(\frac{1}{2}\right)=0 \\ & \Rightarrow \tan ^{-1}\left(\frac{\frac{2 x}{5-x^2}-\frac{1}{2}}{1+\frac{2 x}{5-x^2} \times \frac{1}{2}}\right)=0 \\ & \Rightarrow \tan ^{-1}\left(\frac{x^2+4 x-5}{-x^2+x+5}\right)=0 \\ \therefore \quad & x^2+4 x-5=0 \\ \therefore \quad & x=-5 \text { or } x=1\end{aligned}$

Asked in: MHT CET 2024 (10 May Shift 2)

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