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

We have to prove, $\cot \left( {\frac{\pi }{4} - 2{{\cot }^{ - 1}}3} \right) = 7$

$\Rightarrow \left( {\frac{\pi }{4} - 2{{\cot }^{ - 1}}3} \right) = {\cot ^{ - 1}}7$

$\Rightarrow \left( {2{{\cot }^{ - 1}}3} \right) = \frac{\pi }{4} - {\cot ^{ - 1}}7$

$\Rightarrow$ $2{\tan ^{ - 1}}\frac{1}{3} = \frac{\pi }{4} - {\tan ^{ - 1}}\frac{1}{7}$

$\Rightarrow$ $2{\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{7} = \frac{\pi }{4}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{{2/3}}{{1 - {{(1/3)}^2}}} + {\tan ^{ - 1}}\frac{1}{7} = \frac{\pi }{4}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{{2/3}}{{8/9}} + {\tan ^{ - 1}}\frac{1}{7} = \frac{\pi }{4}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{1}{7} = \frac{\pi }{4}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{{\frac{3}{4} + \frac{1}{7}}}{{1 - \frac{3}{4} \cdot \frac{1}{7}}} = \frac{\pi }{4}$

$\Rightarrow$ ${\tan ^{ - 1}}\frac{{(21 + 4)/28}}{{(28 - 3)/28}} = \frac{\pi }{4}$
$\Rightarrow$ ${\tan ^{ - 1}}\frac{{25}}{{25}} = \frac{\pi }{4}$

$\Rightarrow$ $1 = \tan \frac{\pi }{4}$

$\Rightarrow$ $1 = 1$
$\Rightarrow$ ${\rm{LHS}} = {\rm{RHS}}$