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Given $x=a\left(\cos\theta+\log\tan\frac{\theta}{2}\right)$, differentiate with respect to $\theta$:\r\n$$ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{1}{\tan\frac{\theta}{2}} \cdot \sec^2\frac{\theta}{2} \cdot \frac{1}{2}\right) $$\r\n$$ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{\cos\frac{\theta}{2}}{\sin\frac{\theta}{2}} \cdot \frac{1}{\cos^2\frac{\theta}{2}} \cdot \frac{1}{2}\right) $$\r\n$$ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{1}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}\right) $$\r\n$$ \frac{dx}{d\theta} = a\left(-\sin\theta + \frac{1}{\sin\theta}\right) $$\r\n$$ \frac{dx}{d\theta} = a\left(\frac{1-\sin^2\theta}{\sin\theta}\right) = a\frac{\cos^2\theta}{\sin\theta} $$
Given $y = a\sin\theta$, differentiate with respect to $\theta$:\r\n$$ \frac{dy}{d\theta} = a\cos\theta $$
Using the chain rule, $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$:\r\n$$ \frac{dy}{dx} = \frac{a\cos\theta}{a\frac{\cos^2\theta}{\sin\theta}} = \frac{\sin\theta}{\cos\theta} = \tan\theta $$
Differentiate $\frac{dy}{dx} = \tan\theta$ with respect to $x$:\r\n$$ \frac{d^2y}{dx^2} = \frac{d}{dx}(\tan\theta) = \sec^2\theta \cdot \frac{d\theta}{dx} $$\r\nSince $\frac{dx}{d\theta} = a\frac{\cos^2\theta}{\sin\theta}$, then $\frac{d\theta}{dx} = \frac{\sin\theta}{a\cos^2\theta}$.\r\n$$ \frac{d^2y}{dx^2} = \sec^2\theta \cdot \frac{\sin\theta}{a\cos^2\theta} = \frac{1}{\cos^2\theta} \cdot \frac{\sin\theta}{a\cos^2\theta} = \frac{\sin\theta}{a\cos^4\theta} $$
Substitute $\theta = \frac{\pi}{4}$ into the expression for $\frac{d^2y}{dx^2}$:\r\n$$ \frac{d^2y}{dx^2}\Big|_{\theta=\frac{\pi}{4}} = \frac{\sin\frac{\pi}{4}}{a\cos^4\frac{\pi}{4}} = \frac{\frac{1}{\sqrt{2}}}{a\left(\frac{1}{\sqrt{2}}\right)^4} = \frac{\frac{1}{\sqrt{2}}}{a\left(\frac{1}{4}\right)} = \frac{4}{a\sqrt{2}} = \frac{4\sqrt{2}}{2a} = \frac{2\sqrt{2}}{a} $$
\r\n Final Answer: $\frac{2\sqrt{2}}{a}$<\/span>\r\n <\/p>\r\n <\/div>\r\n <\/div>
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