If x = sin left( 2 tan^{-1} 2 right) and y = | vedclass

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(D) Given $x = \sin(2 	an^{-1} 2)$. Let $	an^{-1} 2 = 	heta$,so $	an 	heta = 2$. Since $	an 	heta = 2$,we have $\sin 	heta = \frac{2}{\sqrt{5}

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$y^2 = 1 - x$

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(D) Given $x = \sin(2 	an^{-1} 2)$. Let $	an^{-1} 2 = 	heta$,so $	an 	heta = 2$. Since $	an 	heta = 2$,we have $\sin 	heta = \frac{2}{\sqrt{5}}$ and $\cos 	heta = \frac{1}{\sqrt{5}}$. Then $x = \sin(2 	heta) = 2 \sin 	heta \cos 	heta = 2 \left( \frac{2}{\sqrt{5}} ight) \left( \frac{1}{\sqrt{5}} ight) = \frac{4}{5}$. Thus,$1 - x = 1 - \frac{4}{5} = \frac{1}{5}$. Now,given $y = \sin \left( \frac{1}{2} 	an^{-1} \frac{4}{3} ight)$. Let $	an^{-1} \frac{4}{3} = \alpha$,so $	an \alpha = \frac{4}{3}$. Using the identity $\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}$,where $\cos \alpha = \frac{3}{5}$ (since $	an \alpha = \frac{4}{3}$),we get $y = \sqrt{\frac{1 - 3/5}{2}} = \sqrt{\frac{2/5}{2}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}}$. Therefore,$y^2 = \frac{1}{5}$. Comparing the results,$y^2 = \frac{1}{5}$ and $1 - x = \frac{1}{5}$,so $y^2 = 1 - x$.

If $x = \sin \left( 2 \tan^{-1} 2 \right)$ and $y = \sin \left( \frac{1}{2} \tan^{-1} \frac{4}{3} \right)$,then -

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