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

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

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$\frac{1}{2}$

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(A) Given $y = 	an^{-1}(\sec x + 	an x)$.We can rewrite the expression inside the inverse tangent function as:$y = 	an^{-1}\left(\frac{1}{\cos x} + \frac{\sin x}{\cos x}ight) = 	an^{-1}\left(\frac{1 + \sin x}{\cos x}ight)$.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:$y = 	an^{-1}\left[\frac{\cos^2\frac{x}{2} + \sin^2\frac{x}{2} + 2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^2\frac{x}{2} - \sin^2\frac{x}{2}}ight]$$y = 	an^{-1}\left[\frac{(\cos\frac{x}{2} + \sin\frac{x}{2})^2}{(\cos\frac{x}{2} - \sin\frac{x}{2})(\cos\frac{x}{2} + \sin\frac{x}{2})}ight]$$y = 	an^{-1}\left[\frac{\cos\frac{x}{2} + \sin\frac{x}{2}}{\cos\frac{x}{2} - \sin\frac{x}{2}}ight]$Dividing numerator and denominator by $\cos\frac{x}{2}$:$y = 	an^{-1}\left(\frac{1 + 	an\frac{x}{2}}{1 - 	an\frac{x}{2}}ight) = 	an^{-1}\left(	an(\frac{\pi}{4} + \frac{x}{2})ight)$$y = \frac{\pi}{4} + \frac{x}{2}$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = 0 + \frac{1}{2} = \frac{1}{2}$.

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

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