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If $x=a(1-\cos \theta), \quad y=a(\theta-\sin \theta)$, then $\frac{d^{2} y}{d x^{2}}=$
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Verified Answer
The correct answer is:
$\frac{\operatorname{cosec} \theta}{2 a \cos ^{2}\left(\frac{\theta}{2}\right)}$
$\frac{d x}{d \theta}=a \sin \theta, \frac{d y}{d \theta}=a(1-\cos \theta)$
$\frac{d y}{d x}=\frac{a(1-\cos \theta)}{a \sin \theta}=\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\tan \frac{\theta}{2}$
$\frac{d^{2} y}{d x^{2}}=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{d \theta}{d x}$
$=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \times \frac{1}{a \sin \theta}=\frac{\operatorname{cosec} \theta}{2 \cos ^{2} \frac{\theta}{2}}$
$\frac{d y}{d x}=\frac{a(1-\cos \theta)}{a \sin \theta}=\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\tan \frac{\theta}{2}$
$\frac{d^{2} y}{d x^{2}}=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{d \theta}{d x}$
$=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \times \frac{1}{a \sin \theta}=\frac{\operatorname{cosec} \theta}{2 \cos ^{2} \frac{\theta}{2}}$
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