Question

Prove that cot (pie/4 - 2 cot inverse 3)=7

Answer 1

$$\begin{gathered} 2{\tan }^{-1}(1/3)={\tan }^{-1}\frac{2\times {\displaystyle \frac{1}{3}}}{1-{\displaystyle \frac{1}{9}}}={\tan }^{-1}(3/4) \\ {\tan }^{-1}(1/7)+{\tan }^{-1}(3/4)={\tan }^{-1}\frac{{\displaystyle \frac{1}{7}}+{\displaystyle \frac{3}{4}}}{1-{\displaystyle \frac{3}{28}}}=\frac{\pi }{4} \\ {\tan }^{-1}(1/7)+{\tan }^{-1}(3/4)=\frac{\pi }{4} \\ {\tan }^{-1}(1/7)=\frac{\pi }{4}-{\tan }^{-1}(3/4) \\ co{t}^{-1}\left(7\right)=\frac{\pi }{4}-2{\tan }^{-1}(1/3) \\ co{t}^{-1}\left(7\right)=\frac{\pi }{4}-2co{t}^{-1}\left(3\right) \\ cot(\frac{\pi }{4}-2co{t}^{-1}(3))=7 \end{gathered}$$