Question

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

Answer 1

Let ${\cot }^{-1}3=A$

$\cot A=3$

$\cot [\frac{\pi }{4}-2{\cot }^{-1}3]=\cot (\frac{\pi }{4}-2A)$

$=\frac{1}{\tan (\frac{\pi }{4}-2A)}=\frac{1+\tan 2A}{1-\tan 2A}....\left(1\right)$

Now, $\cot A=3$

$\tan A=\frac{1}{3}$

Therefore,

$\tan 2A=\frac{2\tan A}{1-{\tan }^{2}A}=\frac{2\times \frac{1}{3}}{1-\frac{1}{9}}=\frac{3}{4}$

$\therefore$ Eq $\left(1\right)$ becomes

$=\frac{1+\frac{3}{4}}{1-\frac{3}{4}}=7=RHS$