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

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Correct option is B

Put tan−12=Aand tan−143=B.⇒tanA=2 and tanB=43 ⇒sinA=25,cosA=15 and cosB=35 ∴x=sin2tan-12=sin2A=2sinAcosA=22515=45 and y=sin12tan-145=sinB2=1-cosB2=1-352=15 ⇒x>y

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If x=sin(2tan−12)and y=sin[12tan−143]then

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