Question

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

• A. $\cot \theta$
• B. $\tan \theta$
• C. $\sin \theta$
• D. $\cos \theta$

Answer 1

The correct option is B $\tan \theta$

$x=a\cos \theta +alog\tan \frac{\theta }{2}$ & $y=a\sin \theta$

$\therefore \frac{dx}{d\theta }=-a\sin \theta +a[\frac{1}{\tan \theta /2}.{\sec }^2\left(\frac{\theta }{2}\right).\frac{1}{2}]$

$\frac{dx}{d\theta }=-a\sin \theta +a\frac{1}{{\cos }^2\theta /2}$

$=-a\sin \theta +\frac{a}{\sin \theta }$

$=a(-\sin \theta +\frac{1}{\sin \theta })$

$\Rightarrow a.\frac{{\cos }^2\theta }{\sin \theta }$ .........$\left(1\right)$

$\frac{dy}{d\theta }=a\frac{d}{d\theta }\left(\sin \theta \right)\Rightarrow a\cos \theta$ ........$\left(2\right)$

$\therefore \frac{dy}{dx}=\frac{\cos \theta }{{\cos }^2\theta }\cdot \sin \theta$

$\frac{dy}{dx}=\tan \theta$.