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

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(C) Let $x = 	an^{-1} \frac{3}{4}$,then $	an x = \frac{3}{4}$. Using $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get $\cos(2 	an^{-1} \frac{3}

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greater than $\frac{\pi }{4}$

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(C) Let $x = 	an^{-1} \frac{3}{4}$,then $	an x = \frac{3}{4}$. Using $\cos 2x = \frac{1 - 	an^2 x}{1 + 	an^2 x}$,we get $\cos(2 	an^{-1} \frac{3}{4}) = \frac{1 - (3/4)^2}{1 + (3/4)^2} = \frac{1 - 9/16}{1 + 9/16} = \frac{7/16}{25/16} = \frac{7}{25}$.Let $y = \cot^{-1} \frac{1}{2}$,then $\cot y = \frac{1}{2}$,so $	an y = 2$. Using $\sin 2y = \frac{2 	an y}{1 + 	an^2 y}$,we get $\sin(2 \cot^{-1} \frac{1}{2}) = \frac{2(2)}{1 + 2^2} = \frac{4}{5} = \frac{20}{25}$.Substituting these values into the expression: $	an^{-1} [\frac{7}{25} + \frac{20}{25}] = 	an^{-1} [\frac{27}{25}]$.Since $\frac{27}{25} > 1$,it follows that $	an^{-1}(\frac{27}{25}) > 	an^{-1}(1)$,which is $\frac{\pi}{4}$.Thus,the value is greater than $\frac{\pi}{4}$.

The value of ${\tan ^{ - 1}}\left[ {\cos \left( {2\,{{\tan }^{ - 1}}\frac{3}{4}} \right)\, + \,\sin \,\left( {2\,{{\cot }^{ - 1}}\frac{1}{2}} \right)} \right]$ is

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