A) \[a\sin t\]
B) \[2a{{\sin }^{3}}(t/2)\sec (t/2)\]
C) \[2a\sin (t/2)\,\,\tan \,(t/2)\]
D) \[2a\sin (t/2)\]
Correct Answer: C
Solution :
\[x=a(t+\sin t)\], \[y=a(1-\cos t)\] \\[\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{a(\sin t)}{a(1+\cos t)}\]=\[\tan \frac{t}{2}\] Length of the normal = \[y\sqrt{1+{{\left( \frac{dy}{dx} \right)}^{2}}}\] = \[a(1-\cos t)\,\sqrt{1+{{\tan }^{2}}(t/2)}\]=\[a(1-\cos t)\,\sec (t/2)\] = \[2a{{\sin }^{2}}(t/2)\,\sec \,(t/2)\]= \[2a\sin (t/2)\,\tan (t/2)\].
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