Show that $\cos \left( {2{{\tan }^{ - 1}}\frac{1}{7}} \right) = \sin \left( {4{{\tan }^{ - 1}}\frac{1}{3}} \right)$.

We have, $\cos \left( {2{{\tan }^{ - 1}}\frac{1}{7}} \right) = \sin \left( {4{{\tan }^{ - 1}}\frac{1}{3}} \right)$

$\Rightarrow$ $\cos \left[ {{{\cos }^{ - 1}}\left( {\frac{{1 - {{\left( {\frac{1}{7}} \right)}^2}}}{{1 + {{\left( {\frac{1}{7}} \right)}^2}}}} \right)} \right] = \sin \left[ {2 \cdot 2{{\tan }^{ - 1}}\frac{1}{3}} \right]$

$\Rightarrow$ $\cos \left[ {{{\cos }^{ - 1}}\left( {\frac{{\frac{{48}}{{49}}}}{{\frac{{50}}{{49}}}}} \right)} \right] = \sin \left[ {2 \cdot \left( {{{\tan }^{ - 1}}\frac{{\frac{2}{3}}}{{1 - {{\left( {\frac{1}{3}} \right)}^2}}}} \right)} \right]$

$\Rightarrow$ $\cos \left[ {{{\cos }^{ - 1}}\left( {\frac{{48 \times 49}}{{50 \times 49}}} \right)} \right] = \sin \left[ {2{{\tan }^{ - 1}}\left( {\frac{{18}}{{24}}} \right)} \right]$

$\Rightarrow$ $\cos \left[ {{{\cos }^{ - 1}}\left( {\frac{{24}}{{25}}} \right)} \right] = \sin \left( {2{{\tan }^{ - 1}}\frac{3}{4}} \right)$

$\Rightarrow$ $\cos \left[ {{{\cos }^{ - 1}}\left( {\frac{{24}}{{25}}} \right)} \right] = \sin \left( {{{\sin }^{ - 1}}\frac{{2 \times \frac{3}{4}}}{{1 + \frac{9}{{16}}}}} \right)$

$\Rightarrow$ $\frac{{24}}{{25}} = \sin \left( {{{\sin }^{ - 1}}\frac{{3/2}}{{25/16}}} \right)$
$\Rightarrow$ $\frac{{24}}{{25}} = \frac{{48}}{{50}} \Rightarrow \frac{{24}}{{25}} = \frac{{24}}{{25}}$
$therefore, {\rm{LHS}} = {\rm{RHS}}$
Hence proved.