If x=tan−1(1)+cos−1(−12)+sin−1(−12) & y=cos(12cos−1(18)) then

cos-1(18)=α⇒y=cos(α2)=1+cosα2=1+182=34 x=π4+2π3−π6=3π4 ⇒x=πy

If x=tan−1(1)+cos−1(−12)+sin−1(−12) & y=cos(12cos−1(18)) then

cos-1(18)=α⇒y=cos(α2)=1+cosα2=1+182=34 x=π4+2π3−π6=3π4 ⇒x=πy

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If $x = \tan^{- 1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2})$ & $y = \cos (\frac{1}{2} \cos^{- 1} (\frac{1}{8}))$ then

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$x = 2 π y$

$y = 3 π x$

$x = π y$

$y = π x$

answer is C.

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${cos}^{-1} (\frac{1}{8}) = α \Rightarrow y = \cos (\frac{α}{2}) = \sqrt{\frac{1 + \cos α}{2}} = \sqrt{\frac{1 + \frac{1}{8}}{2}} = \frac{3}{4} x = \frac{π}{4} + \frac{2 π}{3} - \frac{π}{6} = \frac{3 π}{4} \Rightarrow x = π y$

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If x=tan−1(1)+cos−1(−12)+sin−1(−12) & y=cos(12cos−1(18)) then

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If $x = \tan^{- 1} (1) + \cos^{- 1} (\frac{- 1}{2}) + \sin^{- 1} (\frac{- 1}{2})$ & $y = \cos (\frac{1}{2} \cos^{- 1} (\frac{1}{8}))$ then