Question

cot inverse x -cot inverse(x+2)=pie/12

Answer 1

Dear student

The above equation can be written as

$$\begin{gathered} {\cot }^{-1}\mathrm{x}-{\cot }^{-1}\left(\mathrm{x}+2\right)=\frac{\mathrm{\pi }}{12} \\ \Rightarrow \left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\mathrm{x}\right]-\left[\frac{\mathrm{\pi }}{2}-{\tan }^{-1}\left(\mathrm{x}+2\right)\right]=\frac{\mathrm{\pi }}{12} \\ \Rightarrow {\tan }^{-1}\left(\mathrm{x}+2\right)-{\tan }^{-1}\left(\mathrm{x}\right)=\frac{\mathrm{\pi }}{12} \\ \Rightarrow {\tan }^{-1}\left(\frac{\mathrm{x}+2-\mathrm{x}}{1+\mathrm{x}(\mathrm{x}+2)}\right)=\frac{\mathrm{\pi }}{12} \\ \Rightarrow \frac{2}{{\left(\mathrm{x}+1\right)}^2}=\tan \left(\frac{\mathrm{\pi }}{12}\right) \\ \Rightarrow \frac{2}{{\left(\mathrm{x}+1\right)}^2}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\left[\mathbf{as}\mathbf{,}\mathbf{}\mathbf{tan}\mathbf{}\left(\frac{\mathbf{\pi }}{\mathbf{12}}\right)\mathbf{}\mathbf{=}\mathbf{}\mathbf{tan}\mathbf{}\mathbf{15}\mathbf{°}\mathbf{}\mathbf{=}\mathbf{}\frac{\sqrt{\mathbf{3}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}}{\sqrt{\mathbf{3}}\mathbf{+}\mathbf{1}}\right] \\ \Rightarrow \frac{2}{{\left(\mathrm{x}+1\right)}^2}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1} \\ \Rightarrow \frac{2}{{\left(\mathrm{x}+1\right)}^2}=\frac{3-1}{{\left(\sqrt{3}+1\right)}^2} \\ \Rightarrow \frac{2}{{\left(\mathrm{x}+1\right)}^2}=\frac{2}{{\left(\sqrt{3}+1\right)}^2} \\ \Rightarrow {\left(\mathrm{x}+1\right)}^2={\left(\sqrt{3}+1\right)}^2 \\ \Rightarrow \mathrm{x}=\sqrt{3} \end{gathered}$$

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