Question

Prove that: $\cot \frac{\pi }{8}=\sqrt{2}+1$

Answer 1

$$\begin{gathered} \frac{\pi }{8}={\left(22\frac{{\displaystyle 1}}{{\displaystyle 2}}\right)}^{°} \\ \mathrm{Let}A={\left(22\frac{1}{2}\right)}^{°} \\ \mathrm{Using}\mathrm{the}\mathrm{identity}\cot 2A=\frac{{\cot }^{2}A-1}{2\cot A},\mathrm{we}\mathrm{get} \\ \cot 45°=\frac{{\cot }^2{\left(22\frac{1}{2}\right)}^{°}-1}{2\cot {\left(22\frac{1}{2}\right)}^{°}} \\ \Rightarrow 1=\frac{{\cot }^2{\left(22\frac{1}{2}\right)}^{°}-1}{2\cot {\left(22\frac{1}{2}\right)}^{°}}\left(\because \cot 45°=1\right) \\ \Rightarrow 2\cot {\left(22\frac{1}{2}\right)}^{°}-{\cot }^2{\left(22\frac{1}{2}\right)}^{°}+1=0 \end{gathered}$$

$$\begin{gathered} \Rightarrow {\cot }^2{\left(22\frac{1}{2}\right)}^{\circ }-2\cot {\left(22\frac{1}{2}\right)}^{\circ }-1=0 \\ \Rightarrow \left\{{\cot }^2{\left(22\frac{1}{2}\right)}^{\circ }-2\cot {\left(22\frac{1}{2}\right)}^{\circ }+1\right\}-2=0 \\ \Rightarrow {\left\{\cot {\left(22\frac{1}{2}\right)}^{\circ }-1\right\}}^2=2 \\ \Rightarrow \cot {\left(22\frac{1}{2}\right)}^{\circ }-1=\sqrt{2} \\ \Rightarrow \cot {\left(22\frac{1}{2}\right)}^{\circ }=\sqrt{2}+1 \end{gathered}$$