Question

If $x=a(\cos t+\log \tan \frac{t}{2}),y=a\sin t$, then $\frac{dy}{dx}=$

• A. $\tan t$
• B. $\cot t$
• C. $-\cot t$
• D. $-\tan t$

Answer 1

The correct option is A $\tan t$

Solution:

We have, $x=a(\cos t+\log \tan \frac{t}{2}),y=a\sin t$

Now,

$\frac{dx}{dt}=-a\sin t+\frac{a}{\tan \frac{t}{2}}\times {\sec }^2\frac{t}{2}\times \frac{1}{2}$

$=-a\sin t+\frac{a}{2\sin \frac{t}{2}\cos \frac{t}{2}}$

$=-a\sin t+\frac{a}{\sin t}$ $(\because sint=2\sin \frac{t}{2}\cos \frac{t}{2})$

$=\frac{a(1-{\sin }^{2}t)}{\sin t}$

$=\frac{a{\cos }^{2}t}{\sin t}.................\left(i\right)$

And

$\frac{dy}{dt}=a\cos t................\left(ii\right)$

Now,

From eqn.(i) and eqn.(ii), we get

$\frac{dy}{dt}\times \frac{dt}{dx}=a\cos t\times \frac{\sin t}{a{\cos }^{2}t}$

$⟹\frac{dy}{dx}=\tan t$

Hence, A is the correct option.