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

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(B) Given $y = 	an^{-1}(\sec x - 	an x)$.We can rewrite the expression inside the inverse tangent function as:$\sec x - 	an x = \frac{1}{\cos x} -

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$-\frac{1}{2}$

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(B) Given $y = 	an^{-1}(\sec x - 	an x)$.We can rewrite the expression inside the inverse tangent function as:$\sec x - 	an x = \frac{1}{\cos x} - \frac{\sin x}{\cos x} = \frac{1 - \sin x}{\cos x}$Using the half-angle identities $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$,$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$,and $1 = \cos^2\frac{x}{2} + \sin^2\frac{x}{2}$,we get:$\frac{1 - \sin x}{\cos x} = \frac{(\cos\frac{x}{2} - \sin\frac{x}{2})^2}{(\cos\frac{x}{2} - \sin\frac{x}{2})(\cos\frac{x}{2} + \sin\frac{x}{2})} = \frac{\cos\frac{x}{2} - \sin\frac{x}{2}}{\cos\frac{x}{2} + \sin\frac{x}{2}}$Dividing the 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 $x$,we get:$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} - \frac{x}{2}) = -\frac{1}{2}$

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

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