If y = tan^{-1} sqrt{frac{1 + sin 2x}{1 - sin | vedclass

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

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(B) Given $y = 	an^{-1} \sqrt{\frac{1 + \sin 2x}{1 - \sin 2x}}$.Using $1 = \sin^2 x + \cos^2 x$ and $\sin 2x = 2 \sin x \cos x$, we have $1 + \sin 2x = (\cos x + \sin x)^2$ and $1 - \sin 2x = (\cos x - \sin x)^2$.Thus, $y = 	an^{-1} \sqrt{\frac{(\cos x + \sin x)^2}{(\cos x - \sin x)^2}} = 	an^{-1} \left| \frac{\cos x + \sin x}{\cos x - \sin x} ight|$.Dividing numerator and denominator by $\cos x$, we get $y = 	an^{-1} \left| \frac{1 + 	an x}{1 - 	an x} ight| = 	an^{-1} |	an(\frac{\pi}{4} + x)|$.For $x = \frac{\pi}{6}$, $\frac{\pi}{4} + \frac{\pi}{6} = \frac{5\pi}{12}$, which is in the first quadrant, so $y = \frac{\pi}{4} + x$.Differentiating with respect to $x$, $\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} + x) = 1$.

If $y = \tan^{-1} \sqrt{\frac{1 + \sin 2x}{1 - \sin 2x}}$, then at $x = \frac{\pi}{6}$, the value of $\frac{dy}{dx}$ is

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