If x = sin (2 tan^{-1} 2),y = cos (2 tan^{-1} | vedclass

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(D) We use the identities: $2 	an^{-1} (	heta) = \sin^{-1} (\frac{2 	heta}{1 + 	heta^2}) = \cos^{-1} (\frac{1 - 	heta^2}{1 + 	heta^2}) = \sec^{-

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$z < y < x$

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(D) We use the identities: $2 	an^{-1} (	heta) = \sin^{-1} (\frac{2 	heta}{1 + 	heta^2}) = \cos^{-1} (\frac{1 - 	heta^2}{1 + 	heta^2}) = \sec^{-1} (\frac{1 + 	heta^2}{1 - 	heta^2})$.For $x = \sin (2 	an^{-1} 2)$:$x = \sin (\sin^{-1} (\frac{2 	imes 2}{1 + 2^2})) = \frac{4}{5} = 0.8$.For $y = \cos (2 	an^{-1} 3)$:$y = \cos (\cos^{-1} (\frac{1 - 3^2}{1 + 3^2})) = \frac{1 - 9}{1 + 9} = \frac{-8}{10} = -0.8$.For $z = \sec (2 	an^{-1} 4)$:$z = \sec (\sec^{-1} (\frac{1 + 4^2}{1 - 4^2})) = \frac{1 + 16}{1 - 16} = \frac{17}{-15} \approx -1.133$.Comparing the values: $-1.133 < -0.8 < 0.8$,which means $z < y < x$.

If $x = \sin (2 \tan^{-1} 2)$,$y = \cos (2 \tan^{-1} 3)$,and $z = \sec (2 \tan^{-1} 4)$,then:

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