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

Question

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

Dr. Abubacker · Accepted Answer

(b)
On differentiating given curves w.r.t.$	heta$ respectively, we get
$\frac{dx}{d	heta} = a\left( - \sin	heta + \frac{1}{	an\left( \frac{	heta}{2} ight)}\ .\sec^{2}\frac{	heta}{2}\ .\frac{1}{2} ight)\$
and $\frac{dy}{d	heta}$ = $a\cos	heta$
$\Rightarrow \ \frac{dx}{d	heta}$ = $\frac{a\cos^{2}	heta}{\sin	heta}\ and\ \frac{dy}{d	heta} = a\cos	heta$
$	herefore\ \ \ \frac{dy}{dx} = \frac{dy\ /d	heta}{dx/\ d	heta} = \frac{a\cos	heta}{a\cos^{2}{	heta/\sin	heta}}$= tan$	heta$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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