If y = tan^{-1}(sec x - tan x),then frac{dy}{ | vedclass

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(B) Given $y = 	an^{-1}(\sec x - 	an x)$.First,simplify the expression inside the inverse tangent function:$\sec x - 	an x = \frac{1}{\cos x} - \fr

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$-0.5$

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(B) Given $y = 	an^{-1}(\sec x - 	an x)$.First,simplify the expression inside the inverse tangent function:$\sec x - 	an x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x}$.Using the half-angle identities $1 - \sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = (\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})$:$\frac{1 - \sin x}{\cos x} = \frac{(\cos \frac{x}{2} - \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})} = \frac{\cos \frac{x}{2} - \sin \frac{x}{2}}{\cos \frac{x}{2} + \sin \frac{x}{2}}$.Dividing numerator and denominator by $\cos \frac{x}{2}$,we get $\frac{1 - 	an \frac{x}{2}}{1 + 	an \frac{x}{2}} = 	an(\frac{\pi}{4} - \frac{x}{2})$.Thus,$y = 	an^{-1}(	an(\frac{\pi}{4} - \frac{x}{2})) = \frac{\pi}{4} - \frac{x}{2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} - \frac{x}{2}) = 0 - \frac{1}{2} = -\frac{1}{2} = -0.5$.

If $y = \tan^{-1}(\sec x - \tan x)$,then $\frac{dy}{dx} = $

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