If 3{sin ^{ - 1}}frac{{2x}}{{1 + {x^2}}} - 4{ | vedclass

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(B) Given the equation: $3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} + 2{	an ^{ - 1}}\frac{{2x}}{{1 - {x

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$\frac{1}{{\sqrt 3 }}$

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(B) Given the equation: $3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} + 2{	an ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}$.Substitute $x = 	an 	heta$,which implies $	heta = {	an ^{ - 1}}x$.Using standard trigonometric identities:$\sin 2	heta = \frac{{2	an 	heta }}{{1 + {{	an }^2}	heta }}$,$\cos 2	heta = \frac{{1 - {{	an }^2}	heta }}{{1 + {{	an }^2}	heta }}$,and $	an 2	heta = \frac{{2	an 	heta }}{{1 - {{	an }^2}	heta }}$.The equation becomes:$3(2	heta ) - 4(2	heta ) + 2(2	heta ) = \frac{\pi }{3}$.Simplifying the expression:$6	heta - 8	heta + 4	heta = \frac{\pi }{3}$.$2	heta = \frac{\pi }{3} \Rightarrow 	heta = \frac{\pi }{6}$.Since $	heta = {	an ^{ - 1}}x$,we have ${	an ^{ - 1}}x = \frac{\pi }{6}$.Therefore,$x = 	an \frac{\pi }{6} = \frac{1}{{\sqrt 3 }}$.

If $3{\sin ^{ - 1}}\frac{{2x}}{{1 + {x^2}}} - 4{\cos ^{ - 1}}\frac{{1 - {x^2}}}{{1 + {x^2}}} + 2{\tan ^{ - 1}}\frac{{2x}}{{1 - {x^2}}} = \frac{\pi }{3}$,then $x$ =

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