Question

Prove that $\cot (\frac{\pi }{4}-2{\cot }^{-1}3)=7$.

Answer 1

$cot(\frac{\pi }{4}-2co{t}^{-1}3)=7let2{\cot }^{-1}3=\theta {\cot }^{-1}3=\frac{\theta }{2}\cot \frac{\theta }{2}=3\tan \frac{\theta }{2}=\frac{1}{3}\Rightarrow \cot \left(\frac{\pi }{4}-\theta \right)=7\Rightarrow \frac{\cot \frac{\pi }{4}\cdot \cot \theta +1}{\cot \theta -\cot \frac{\pi }{4}}=7\Rightarrow \frac{\cot \theta +1}{\cot \theta -1}=7\Rightarrow \frac{1+\tan \theta }{1-\tan \theta }=\frac{\left(1+\frac{2\tan \frac{\theta }{2}}{1\cdot {\tan }^2\frac{\theta }{2}}\right)}{\left(1-\frac{2\tan \frac{\theta }{2}}{1\cdot {\tan }^2\frac{\theta }{2}}\right)}Put\tan \frac{\theta }{2}=\frac{1}{3}\Rightarrow \frac{\left(1+\frac{\frac{2}{3}}{1-\frac{1}{9}}\right)}{\left(1-\frac{\frac{2}{3}}{1-\frac{1}{9}}\right)}\Rightarrow \frac{1+\frac{2}{3}\times \frac{9}{8}}{1-\frac{2}{3}\times \frac{9}{8}}\Rightarrow \frac{1+\frac{3}{4}}{1-\frac{3}{4}}\Rightarrow \frac{\frac{7}{4}}{\frac{1}{4}}=7\therefore \cot \left(\frac{\pi }{4}-2{\cot }^{-1}3\right)=7proved$