If y = tan^{-1}(sec x^3 - tan x^3) and frac{p | vedclass

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(B) Given $y = 	an^{-1}(\sec x^3 - 	an x^3)$.$y = 	an^{-1}\left(\frac{1 - \sin x^3}{\cos x^3}ight) = 	an^{-1}\left(\frac{1 - \cos(\frac{\pi}{2}

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$x^2 y'' - 6 y + \frac{3\pi}{2} = 0$

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(B) Given $y = 	an^{-1}(\sec x^3 - 	an x^3)$.$y = 	an^{-1}\left(\frac{1 - \sin x^3}{\cos x^3}ight) = 	an^{-1}\left(\frac{1 - \cos(\frac{\pi}{2} - x^3)}{\sin(\frac{\pi}{2} - x^3)}ight)$.Using the half-angle formulas,$1 - \cos 	heta = 2 \sin^2(\frac{	heta}{2})$ and $\sin 	heta = 2 \sin(\frac{	heta}{2}) \cos(\frac{	heta}{2})$,we get:$y = 	an^{-1}\left(	an(\frac{\pi}{4} - \frac{x^3}{2})ight)$.Since $\frac{\pi}{2} Adding $\frac{\pi}{4}$,we get $-\frac{\pi}{2} Thus,$y = \frac{\pi}{4} - \frac{x^3}{2}$.Now,differentiate with respect to $x$:$y' = -\frac{1}{2} \cdot 3x^2 = -\frac{3}{2}x^2$.$y'' = -\frac{3}{2} \cdot 2x = -3x$.From $y = \frac{\pi}{4} - \frac{x^3}{2}$,we have $x^3 = \frac{\pi}{2} - 2y$.Also,$x^2 = -\frac{2}{3} y'$.Substituting $x^2$ into $y'' = -3x$,we get $y'' = -3 \sqrt[3]{-\frac{2}{3} y'}$.Alternatively,using $x^2 = -\frac{2}{3} y'$,we have $x^2 y'' = x^2 (-3x) = -3x^3 = -3(\frac{\pi}{2} - 2y) = -\frac{3\pi}{2} + 6y$.Therefore,$x^2 y'' - 6y + \frac{3\pi}{2} = 0$.

If $y = \tan^{-1}(\sec x^3 - \tan x^3)$ and $\frac{\pi}{2} < x^3 < \frac{3\pi}{2}$,then:

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