Question

Solve this:
$$\begin{gathered} 40.\mathrm{If}{\tan }^{-1}\frac{\sqrt{1+{\mathrm{x}}^2}-1}{\mathrm{x}}=\frac{\mathrm{\pi }}{45},\mathrm{then} \\ \left(1\right)\mathrm{x}=\tan 2°\left(2\right)\mathrm{x}=\tan 4° \\ \left(3\right)\mathrm{x}=\tan {\left(\frac{1}{4}\right)}^{°}\left(4\right)\mathrm{x}=\tan 8° \end{gathered}$$

Answer 1

$$\begin{gathered} {\tan }^{-1}\frac{\sqrt{1+x^2}-1}{x}=\frac{\mathrm{\pi }}{45} \\ \frac{\sqrt{1+x^2}-1}{x}=\tan \frac{\mathrm{\pi }}{45} \\ \frac{\mathrm{\pi }}{45}=\frac{\mathrm{\pi }}{45}\times \frac{180°}{\mathrm{\pi }}=4° \\ \frac{\sqrt{1+x^2}-1}{x}=\tan 4° \\ Putx=\tan y \\ \frac{\sqrt{1+{\tan }^{2}y}-1}{\tan y}=\tan 4° \\ \frac{\sqrt{se{c}^{2}y}-1}{\tan y}=\tan 4° \\ \frac{secy-1}{\tan y}=\tan 4° \\ \frac{{\displaystyle \frac{1}{\cos y}}-1}{\frac{\sin y}{\cos y}}=\tan 4° \\ \frac{{\displaystyle \frac{1-\cos y}{\cos y}}}{\frac{\sin y}{\cos y}}=\tan 4° \\ \frac{{\displaystyle 1-\cos y}}{\sin y}=\tan 4° \\ \frac{{\displaystyle 2{\sin }^2\frac{y}{2}}}{2\sin {\displaystyle \frac{y}{2}}\cos \frac{y}{2}}=\tan 4° \\ \frac{{\displaystyle \sin \frac{y}{2}}}{\cos \frac{y}{2}}=\tan 4° \\ \tan \frac{y}{2}=\tan 4° \\ \frac{y}{2}=4° \\ y=8° \\ x=\tan y=\tan 8° \end{gathered}$$