frac{d}{dx} tan^{-1} left[ frac{cos x - sin x | vedclass

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

Vedclass · Accepted Answer

$-1$

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(D) Let $y = 	an^{-1} \left[ \frac{\cos x - \sin x}{\cos x + \sin x} ight]$.Divide the numerator and denominator by $\cos x$:$y = 	an^{-1} \left[ \frac{1 - 	an x}{1 + 	an x} ight]$.Using the trigonometric identity $	an(\frac{\pi}{4} - x) = \frac{	an(\frac{\pi}{4}) - 	an x}{1 + 	an(\frac{\pi}{4}) 	an x} = \frac{1 - 	an x}{1 + 	an x}$:$y = 	an^{-1} [	an(\frac{\pi}{4} - x)]$.Since $	an^{-1}(	an 	heta) = 	heta$ for the appropriate range:$y = \frac{\pi}{4} - x$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx} (\frac{\pi}{4} - x) = 0 - 1 = -1$.

$\frac{d}{dx} \tan^{-1} \left[ \frac{\cos x - \sin x}{\cos x + \sin x} \right] = $

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