frac{d}{dx} sqrt{frac{1 - sin 2x}{1 + sin 2x} | vedclass

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Question

(B) Let $y = \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}}$.Using the identities $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:$y = \sqrt

Vedclass · Accepted Answer

$-\sec^2 \left( \frac{\pi}{4} - x \right)$

Vedclass · Answer

(B) Let $y = \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}}$.Using the identities $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$,we have:$y = \sqrt{\frac{\cos^2 x + \sin^2 x - 2 \sin x \cos x}{\cos^2 x + \sin^2 x + 2 \sin x \cos x}} = \sqrt{\frac{(\cos x - \sin x)^2}{(\cos x + \sin x)^2}} = \frac{\cos x - \sin x}{\cos x + \sin x}$.Dividing numerator and denominator by $\cos x$:$y = \frac{1 - 	an x}{1 + 	an x} = 	an \left( \frac{\pi}{4} - x ight)$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = \sec^2 \left( \frac{\pi}{4} - x ight) \cdot \frac{d}{dx} \left( \frac{\pi}{4} - x ight) = \sec^2 \left( \frac{\pi}{4} - x ight) \cdot (-1) = -\sec^2 \left( \frac{\pi}{4} - x ight)$.

$\frac{d}{dx} \sqrt{\frac{1 - \sin 2x}{1 + \sin 2x}} = $

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