If y = tan^{-1}left(frac{sin x + cos x}{cos x | vedclass

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(D) Given that,$y = 	an^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}ight)$.Dividing the numerator and denominator by $\cos x$,we get:$y = 	an

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$1$

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(D) Given that,$y = 	an^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}ight)$.Dividing the numerator and denominator by $\cos x$,we get:$y = 	an^{-1}\left(\frac{	an x + 1}{1 - 	an x}ight)$.Using the formula $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$,where $A = \pi/4$ and $B = x$:$y = 	an^{-1}\left(	an\left(\frac{\pi}{4} + xight)ight) = \frac{\pi}{4} + x$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + xight) = 0 + 1 = 1$.

If $y = \tan^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}\right)$,then $\frac{dy}{dx}$ is equal to

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