Question

what is the value of cot 180/8 and tan 180/8?

Answer 1

$$\begin{gathered} \cot \left(\frac{{180}^{°}}{8}\right)=\cot \frac{\mathrm{\pi }}{8} \\ \mathrm{Now},\cot \frac{\mathrm{\pi }}{8}=\frac{\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}} \\ =\frac{\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}\times \frac{2\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{2\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}\left[\mathbf{multiplying}\mathbf{}\mathbf{numerator}\mathbf{}\mathbf{and}\mathbf{}\mathbf{denominator}\mathbf{}\mathbf{by}\mathbf{}\mathbf{2}\mathbf{}\mathbf{cos}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{8}}\right] \\ =\frac{2{\cos }^2{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{2\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}} \\ =\frac{1+\cos \left(2\times {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)}{\sin \left(2\times {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)}\mathbf{}\mathbf{}\left[\begin{array}{c}\mathbf{2}\mathbf{}{\mathbf{cos}}^{\mathbf{2}}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{\theta }\\ \mathbf{2}\mathbf{}\mathbf{sin\theta }\mathbf{}\mathbf{cos\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{\theta }\end{array}\right] \\ =\frac{1+\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}}{\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}} \\ =\frac{1+{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}} \\ =\frac{{\displaystyle \frac{\sqrt{2}+1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}}=\frac{\sqrt{2}+1}{\sqrt{2}}\times \frac{\sqrt{2}}{1}=\sqrt{2}+1 \end{gathered}$$



$$\begin{gathered} \tan \left(\frac{{180}^{°}}{8}\right)=\tan \frac{\mathrm{\pi }}{8} \\ \mathrm{Now}\tan \frac{\mathrm{\pi }}{8}=\frac{\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}} \\ =\frac{\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}\times \frac{2\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{2\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}\mathbf{}\left[\mathbf{multiplying}\mathbf{}\mathbf{numerator}\mathbf{}\mathbf{and}\mathbf{}\mathbf{denominator}\mathbf{}\mathbf{by}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\mathbf{}\frac{\mathbf{\pi }}{\mathbf{8}}\right] \\ =\frac{2{\sin }^2{\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}}{2\sin {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\cos {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}} \\ =\frac{1-\cos \left(2\times {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)}{\sin \left(2\times {\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$8$}\right.}\right)}\mathbf{}\mathbf{}\mathbf{}\left[\begin{array}{c}\mathbf{as}\mathbf{}\mathbf{2}\mathbf{}{\mathbf{sin}}^{\mathbf{2}}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\mathbf{}\mathbf{-}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{\theta }\\ \mathbf{and}\mathbf{}\mathbf{2}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\theta }\mathbf{}\mathbf{=}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{2}\mathbf{\theta }\end{array}\right] \\ =\frac{1-\cos {\displaystyle \frac{\mathrm{\pi }}{4}}}{\sin {\displaystyle \frac{\mathrm{\pi }}{4}}} \\ =\frac{1-{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}}=\frac{{\displaystyle \frac{\sqrt{2}-1}{\sqrt{2}}}}{{\displaystyle \frac{1}{\sqrt{2}}}}=\frac{\sqrt{2}-1}{\sqrt{2}}\times \frac{\sqrt{2}}{1}=\sqrt{2}-1 \end{gathered}$$