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 $\tan \theta$

$\frac{dx}{d\theta }=\frac{d}{d\theta }(a\cos \left(\theta \right)+a\ln \left(\tan \left(\frac{\theta }{2}\right)\right))$ $=\frac{d}{d\theta }\left(a\cos \left(\theta \right)\right)+a\frac{d}{d\theta }\left(\ln \left(\tan \left(\frac{\theta }{2}\right)\right)\right)$

$=-a\sin \left(\theta \right)+a\csc \left(\theta \right)$

$\frac{dy}{d\theta }=\frac{d}{d\theta }\left(a\sin \left(\theta \right)\right)$ $=a\frac{d}{d\theta }\left(\sin \left(\theta \right)\right)$

$=a\cos \left(\theta \right)$

$\frac{dy}{dx}=\frac{dy}{d\theta }\times \frac{d\theta }{dx}=\frac{acos\theta }{\frac{-asi{n}^2\theta +a}{sin\theta }}=tan\theta$