Question

If $x=\sin t,y=t\cos t.$ Then $\frac{dy}{dx}$ is equal to

• A. $1+t\tan t$
• B. $1-t\tan t$
• C. $1-\tan t$
• D. $\tan t-t\tan t$

Answer 1

The correct option is B

$1-t\tan t$

Here $x$ and $y$ are represented parametrically in terms of $t$.

$\therefore \frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{dy}{dx}$

$\Rightarrow \frac{dy}{dt}=\frac{d\left(t\cos t\right)}{dt}=t(-\sin t)+\cos t$

$\Rightarrow \frac{dx}{dt}=\frac{d\left(\sin t\right)}{dt}=\cos t$

$\Rightarrow \frac{dy}{dx}=\frac{\cos t-t\sin t}{\cos t}=1-t\tan t$

$\therefore \frac{dy}{dx}=1-t\tan t$