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$x=a(1-\sin t), y=a(1-\tan t)$, then $\frac{d y}{d x}$ is |
$\sec ^2 t$ $\sec ^3 t$ $\cos ^2 t$ $\cos ^3 t$ |
$\sec ^3 t$ |
$x=a(1-\sin t) \Rightarrow \frac{d x}{d t}=a(0-\cos t)=-a \cos t$ $\frac{d y}{d t}=\frac{-a \sec ^2 t}{-a \cos t}=\sec ^3 t$ Hence (2) is correct answer. |
