If x=aleft{cos theta+log tan left(frac{theta} | vedclass

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(B) Given that, $x=a\left(\cos 	heta+\log 	an \left(\frac{	heta}{2}ight)ight)$ and $y=a \sin 	heta$. On differentiating $x$ and $y$ with respe

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$	an 	heta$

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(B) Given that, $x=a\left(\cos 	heta+\log 	an \left(\frac{	heta}{2}ight)ight)$ and $y=a \sin 	heta$. On differentiating $x$ and $y$ with respect to $	heta$, we get: $\frac{dx}{d	heta} = a \left( -\sin 	heta + \frac{1}{	an(	heta/2)} \cdot \sec^2(	heta/2) \cdot \frac{1}{2} ight)$ $= a \left( -\sin 	heta + \frac{\cos(	heta/2)}{\sin(	heta/2)} \cdot \frac{1}{\cos^2(	heta/2)} \cdot \frac{1}{2} ight)$ $= a \left( -\sin 	heta + \frac{1}{2 \sin(	heta/2) \cos(	heta/2)} ight) = a \left( -\sin 	heta + \frac{1}{\sin 	heta} ight)$ $= a \left( \frac{1 - \sin^2 	heta}{\sin 	heta} ight) = \frac{a \cos^2 	heta}{\sin 	heta}$. Also, $\frac{dy}{d	heta} = a \cos 	heta$. Therefore, $\frac{dy}{dx} = \frac{dy/d	heta}{dx/d	heta} = \frac{a \cos 	heta}{a \cos^2 	heta / \sin 	heta} = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$.

If $x=a\left\{\cos \theta+\log \tan \left(\frac{\theta}{2}\right)\right\}$ and $y=a \sin \theta$, then $\frac{dy}{dx}$ is equal to

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