Question

The angle made by the tangent at any point on the curve $x=a(t+\sin t\cos t),y=a{(1+\sin t)}^2$ with x-axis is

• A. $\frac{\pi }{2}$
• B. $\frac{\pi }{4}$
• C. $\pi +\frac{t}{2}$
• D. $\frac{\pi }{4}+\frac{t}{2}$

Answer 1

The correct option is D $\frac{\pi }{4}+\frac{t}{2}$

Given,

$x=a(t+\sin t\cos t)$

$\frac{dx}{dt}=\frac{d}{dt}a(t+\sin t\cos t)$

$\therefore \frac{dx}{dt}=a[1+{\cos }^{2}t-{\sin }^{2}t]=a(1+\cos \left(2t\right))=2a{\cos }^{2}t$

$y=a(1+\sin t{)}^2$

$\frac{dy}{dt}=\frac{d}{dt}a(1+\sin t{)}^2$

$\therefore \frac{dy}{dt}=2a(1+\sin \left(t\right))\cos \left(t\right)$

$\Rightarrow \frac{dy}{dx}$

$=\frac{2a(1+\sin \left(t\right))\cos \left(t\right)}{2a{\cos }^{2}t}$

$=\frac{1+\sin t}{\cos t}$

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

$=\frac{1+\tan \frac{t}{2}}{1-\tan \frac{t}{2}}$

$=\tan (\frac{\pi }{4}+\frac{t}{2})$

angle made by tangent,

$\tan \theta =\tan (\frac{\pi }{4}+\frac{t}{2})$

$\Rightarrow \theta =\frac{\pi }{4}+\frac{t}{2}$