Question

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:

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

Answer 1

The correct option is B

$\tan \theta$

Explanation for the correct option.

Step 1: Differentiate $x$ w.r.t $\theta$.

$\begin{array}{rcl}x& =& a\left\{\cos \theta +\log \tan \left(\frac{\theta }{2}\right)\right\}\\ & \Rightarrow & \frac{dx}{d\theta }=a\left\{-\sin \theta +\frac{1}{\tan \left(\frac{\theta }{2}\right)}\times se{c}^2\frac{\theta }{2}\times \frac{1}{2}\right\}\\ & =& a\left\{-\sin \theta +\frac{{\displaystyle \frac{1}{{\cos }^2{\displaystyle \frac{\theta }{2}}}}}{{\displaystyle \frac{\sin {\displaystyle \frac{\theta }{2}}}{\cos {\displaystyle \frac{\theta }{2}}}}}\times \frac{1}{2}\right\}\\ & =& a\left[-\sin \theta +\frac{1}{2\begin{array}{l}\sin \end{array}\begin{array}{l}\frac{\theta }{2}\end{array}\begin{array}{l}\cos \end{array}\begin{array}{l}\frac{\theta }{2}\end{array}}\right]\\ & =& a\left[-\sin \theta +\frac{1}{\sin \theta }\right]\end{array}$

Step 2: Differentiate $y$ w.r.t $\theta$.

$\begin{array}{rcl}y& =& a\sin \theta \\ & \Rightarrow & \frac{dy}{d\theta }=a\cos \theta \end{array}$

Step 3: Find $\frac{dy}{dx}$ .

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{d\theta }\times \frac{d\theta }{dx}\\ & =& \frac{a\cos \theta }{a\left\{-\sin \theta +{\displaystyle \frac{1}{\sin \theta }}\right\}}\\ & =& \frac{\cos \theta }{\left\{{\displaystyle \frac{1-{\sin }^2\theta }{\sin \theta }}\right\}}\\ & =& \frac{\cos \theta }{\left\{{\displaystyle \frac{1-{\sin }^2\theta }{\sin \theta }}\right\}}\\ & =& \frac{\cos \theta }{\left\{{\displaystyle \frac{{\cos }^2\theta }{\sin \theta }}\right\}}\\ & =& \tan \theta \end{array}$

Hence, option B is correct.