If y=log sqrt{frac{1+sin x}{1-sin x}},then fr | vedclass

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(A) Given $y = \log \sqrt{\frac{1+\sin x}{1-\sin x}} = \frac{1}{2} \log \left( \frac{1+\sin x}{1-\sin x} \right) = \frac{1}{2} \{ \log(1+\sin x) - \lo

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$2$

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(A) Given $y = \log \sqrt{\frac{1+\sin x}{1-\sin x}} = \frac{1}{2} \log \left( \frac{1+\sin x}{1-\sin x} \right) = \frac{1}{2} \{ \log(1+\sin x) - \log(1-\sin x) \}$.Now,differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{2} \left( \frac{\cos x}{1+\sin x} - \frac{-\cos x}{1-\sin x} \right) = \frac{1}{2} \cos x \left( \frac{1}{1+\sin x} + \frac{1}{1-\sin x} \right)$.Simplifying the expression inside the bracket:$\frac{dy}{dx} = \frac{1}{2} \cos x \left( \frac{1-\sin x + 1+\sin x}{(1+\sin x)(1-\sin x)} \right) = \frac{1}{2} \cos x \left( \frac{2}{1-\sin^2 x} \right) = \frac{1}{2} \cos x \left( \frac{2}{\cos^2 x} \right) = \frac{1}{\cos x} = \sec x$.At $x = \frac{\pi}{3}$,$\frac{dy}{dx} = \sec \left( \frac{\pi}{3} \right) = 2$.

If $y=\log \sqrt{\frac{1+\sin x}{1-\sin x}}$,then $\frac{d y}{d x}$ at $x=\frac{\pi}{3}$ is

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