The value of sin left( {2{{tan }^{ - 1}}left( | vedclass

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(B) Let the expression be $E = \sin \left( {2{{	an }^{ - 1}}\left( {\frac{1}{3}} ight)} ight) + \cos ({	an ^{ - 1}}(2\sqrt 2 ))$. First,use the

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$\frac{14}{15}$

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(B) Let the expression be $E = \sin \left( {2{{	an }^{ - 1}}\left( {\frac{1}{3}} ight)} ight) + \cos ({	an ^{ - 1}}(2\sqrt 2 ))$. First,use the formula $2	an^{-1}(x) = 	an^{-1}\left(\frac{2x}{1-x^2}ight)$. For $x = \frac{1}{3}$,we have $2	an^{-1}\left(\frac{1}{3}ight) = 	an^{-1}\left(\frac{2/3}{1-1/9}ight) = 	an^{-1}\left(\frac{2/3}{8/9}ight) = 	an^{-1}\left(\frac{3}{4}ight)$. Now,$\sin(	an^{-1}(3/4))$: If $	an(	heta) = 3/4$,then $\sin(	heta) = 3/5$. Next,for $\cos(	an^{-1}(2\sqrt{2}))$: If $	an(\phi) = 2\sqrt{2}$,then $\sec^2(\phi) = 1 + 	an^2(\phi) = 1 + (2\sqrt{2})^2 = 1 + 8 = 9$. So,$\sec(\phi) = 3$,which means $\cos(\phi) = 1/3$. Adding these values: $E = \frac{3}{5} + \frac{1}{3} = \frac{9+5}{15} = \frac{14}{15}$.

The value of $\sin \left( {2{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)} \right) + \cos ({\tan ^{ - 1}}(2\sqrt 2 ))$ is:

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