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

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(D) Given $y = 	an^{-1} \left[ \frac{\sin x + \cos x}{\cos x - \sin x} ight]$.Divide the numerator and denominator by $\cos x$:$y = 	an^{-1} \left

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$1$

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

If $y = \tan^{-1} \left[ \frac{\sin x + \cos x}{\cos x - \sin x} \right]$,then $\frac{dy}{dx}$ is

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