frac{1}{2}{cos ^{ - 1}}left( {frac{{1 - x}}{{ | vedclass

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(B) Let $x = {	an ^2}	heta$,which implies $\sqrt{x} = 	an 	heta$,so $	heta = {	an ^{ - 1}}\sqrt x $.Substituting $x = {	an ^2}	heta$ into the

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${	an ^{ - 1}}\sqrt x $

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(B) Let $x = {	an ^2}	heta$,which implies $\sqrt{x} = 	an 	heta$,so $	heta = {	an ^{ - 1}}\sqrt x $.Substituting $x = {	an ^2}	heta$ into the expression:$\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} ight) = \frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}} ight)$Using the trigonometric identity $\cos 2	heta = \frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}$:$= \frac{1}{2}{\cos ^{ - 1}}(\cos 2	heta)$$= \frac{1}{2}(2	heta) = 	heta$Substituting back $	heta = {	an ^{ - 1}}\sqrt x $:$= {	an ^{ - 1}}\sqrt x $.Thus,the correct option is $B$.

$\frac{1}{2}{\cos ^{ - 1}}\left( {\frac{{1 - x}}{{1 + x}}} \right) = $

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