Question

write the following in the simplest form tan^-1 (x+square root of (1+x^2)) , x belongs to R

Answer 1

$$\begin{gathered} Towrite{\tan }^{-1}\left(x+\sqrt{x^2+1}\right)intothesimplestform,letx=\tan A\Rightarrow A={\tan }^{-1}xSo, \\ {\tan }^{-1}\left(x+\sqrt{x^2+1}\right)={\tan }^{-1}\left(\tan A+\sqrt{{\tan }^{2}A+1}\right) \\ ={\tan }^{-1}\left(\tan A+\sqrt{se{c}^{2}A}\right) \\ ={\tan }^{-1}\left(\tan A+secA\right) \\ ={\tan }^{-1}\left(\frac{\sin A}{\cos A}+\frac{1}{\cos A}\right) \\ ={\tan }^{-1}\left(\frac{\sin A+1}{\cos A}\right) \\ ={\tan }^{-1}\left(\frac{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-A\right)+1}{\sin \left(\frac{\mathrm{\pi }}{2}-A\right)}\right) \\ ={\tan }^{-1}\left(\frac{2{\cos }^2\left({\displaystyle \frac{\mathrm{\pi }}{4}}-{\displaystyle \frac{A}{2}}\right)-1+1}{2\sin \left(\frac{\mathrm{\pi }}{4}-\frac{A}{2}\right)\cos \left(\frac{\mathrm{\pi }}{4}-\frac{A}{2}\right)}\right) \\ ={\tan }^{-1}\left(\frac{\cos \left({\displaystyle \frac{\mathrm{\pi }}{4}}-{\displaystyle \frac{A}{2}}\right)}{\sin \left(\frac{\mathrm{\pi }}{4}-\frac{A}{2}\right)}\right) \\ ={\tan }^{-1}\left(cot\left(\frac{\mathrm{\pi }}{4}-\frac{A}{2}\right)\right) \\ ={\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{2}-\left(\frac{\mathrm{\pi }}{4}-\frac{A}{2}\right)\right)\right) \\ ={\tan }^{-1}\left(\tan \left(\frac{\mathrm{\pi }}{4}+\frac{A}{2}\right)\right) \\ =\left(\frac{\mathrm{\pi }}{4}+\frac{A}{2}\right) \\ =\left(\frac{\mathrm{\pi }}{4}+\frac{{\tan }^{-1}x}{2}\right) \\ Hence,\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}\mathbf{1}}\left(\mathbf{x}\mathbf{+}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}}\right)\mathbf{=}\mathbf{=}\left(\frac{\mathbf{\pi }}{\mathbf{4}}+\frac{\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{x}\mathbf{}}{\mathbf{2}}\right) \end{gathered}$$