Find the value of $\sin \left( {2{{\tan }^{ – 1}}\frac{1}{3}} \right) + \cos \left( {{{\tan }^{ – 1}}2\sqrt 2 } \right)$. [NCERT,Exemplar.2.3,Q.8,Page.36]

We have, $\sin \left( {2{{\tan }^{ – 1}}\frac{1}{3}} \right) + \cos \left( {{{\tan }^{ – 1}}2\sqrt 2 } \right)$

$ = \sin \left[ {{{\sin }^{ – 1}}\left\{ {\frac{{2 \times \frac{1}{3}}}{{1 + {{\left( {\frac{1}{3}} \right)}^2}}}} \right\}} \right] + \cos \left( {{{\cos }^{ – 1}}\frac{1}{3}} \right)$

and $\left. {{{\tan }^{ – 1}}(2\sqrt 2 ) = {{\cos }^{ – 1}}\frac{1}{3}} \right]$

$ = \sin \left[ {{{\sin }^{ – 1}}\left( {\frac{{\frac{2}{3}}}{{1 + \frac{1}{9}}}} \right)} \right] + \frac{1}{3}$

$ = \sin \left[ {{{\sin }^{ – 1}}\left( {\frac{{2 \times 9}}{{3 \times 10}}} \right)} \right] + \frac{1}{3} = \sin \left[ {{{\sin }^{ – 1}}\left( {\frac{3}{5}} \right)} \right] + \frac{1}{3}$

$ = \frac{3}{5} + \frac{1}{3} = \frac{{9 + 5}}{{15}} = \frac{{14}}{{15}}$


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