Question

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

• A. $\frac{1}{4}(\pi +2t)$
• B. $\frac{1-\sin t}{\cos t}$
• C. $\frac{1}{4}(2t-\pi )$
• D. $\frac{1+\sin t}{\cos 2t}$

Answer 1

Answer: (A)

$\frac{dx}{dt}=a+\frac{a}{2}2\cos 2t=a[1+\cos 2t)=2a{\cos }^{2}t$
$\frac{dy}{dt}=2a(1+\sin t)\cdot \cos t$
$\frac{dy}{dx}=\frac{2a(1+\sin t)\cdot \cos t}{2a{\cos }^{2}t}=\frac{(1+\sin t)}{\cos t}$
$\tan \theta =\frac{(\cos (t/2)+\sin (t/2){)}^2}{{\cos }^2(t/2)-{\sin }^2(t/2)}=\frac{1+\tan \frac{t}{2}}{1-\tan \frac{t}{2}}=\tan \left(\frac{\pi }{4}+\frac{t}{2}\right)\Rightarrow \theta =\frac{\pi +2t}{4}$