Pls answer fast MATHS : INVERSE TRIGONOMETRY 

Pls answer fast MATHS : INVERSE TRIGONOMETRY  18.

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{We} \mathrm{have} \\ {\tan }^{-1}\left(\mathrm{x}+1\right)+{\tan }^{-1}\left(\mathrm{x}-1\right)={\cot }^{-1}\left(\frac{1}{2}\right) \\ \mathrm{Use} {\tan }^{-1}\mathrm{A}+{\tan }^{-1}\mathrm{B}={\tan }^{-1}\left(\frac{\mathrm{A}+\mathrm{B}}{1-\mathrm{AB}}\right) \\ \Rightarrow {\tan }^{-1}\left\{\frac{\mathrm{x}+1+\mathrm{x}-1}{1-\left(\mathrm{x}+1\right)\left(\mathrm{x}-1\right)}\right\}={\cot }^{-1}\left(\frac{1}{2}\right) \\ \Rightarrow {\tan }^{-1}\left\{\frac{\mathrm{x}+1+\mathrm{x}-1}{1-\left(\mathrm{x}+1\right)\left(\mathrm{x}-1\right)}\right\}={\tan }^{-1}\left(2\right) \left[{\cot }^{-1}\mathrm{x}={\tan }^{-1}\frac{1}{\mathrm{x}}\right] \\ \Rightarrow {\tan }^{-1}\left\{\frac{2\mathrm{x}}{1-\left({\mathrm{x}}^2-1\right)}\right\}={\tan }^{-1}\left(2\right) \\ \Rightarrow {\tan }^{-1}\left\{\frac{2\mathrm{x}}{1-{\mathrm{x}}^2+1}\right\}={\tan }^{-1}\left(2\right) \\ \Rightarrow {\tan }^{-1}\left\{\frac{2\mathrm{x}}{2-{\mathrm{x}}^2}\right\}={\tan }^{-1}\left(2\right) \\ \Rightarrow \frac{2\mathrm{x}}{2-{\mathrm{x}}^2}=2 \\ \Rightarrow \frac{\mathrm{x}}{2-{\mathrm{x}}^2}=1 \\ \Rightarrow \mathrm{x}=2-{\mathrm{x}}^2 \\ \Rightarrow {\mathrm{x}}^2+\mathrm{x}-2=0 \\ \Rightarrow {\mathrm{x}}^2+2\mathrm{x}-\mathrm{x}-2=0 \\ \Rightarrow \mathrm{x}\left(\mathrm{x}+2\right)-1\left(\mathrm{x}+2\right)=0 \\ \Rightarrow \left(\mathrm{x}-1\right)\left(\mathrm{x}+2\right)=0 \\ \Rightarrow \mathrm{x}=1 \mathrm{or} \mathrm{x}=-2 \\ \mathrm{But} \mathrm{x}=-2 \mathrm{will} \mathrm{not} \mathrm{satisfy} \mathrm{original} \mathrm{equation}, \mathrm{hence} \\ \mathrm{x}=1 \left[\mathrm{Answer}\right] \end{gathered}$$

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