Question

The angle made by the tangent of the curve $x=a(t+\sin t\cos t);y=a(1+\sin t{)}^2$ 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

The correct option is A $\frac{1}{4}(\pi +2t)$

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

$\frac{dx}{dt}=a(1+{\cos }^{2}t-{\sin }^{2}t)$

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

$\frac{dx}{dt}=a(1+\cos 2t)$

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

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

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

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

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

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

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

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

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

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

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

$=\frac{\tan \frac{\pi }{4}+\tan \frac{t}{2}}{1-\tan \frac{\pi }{4}\tan \frac{t}{2}}$

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

If $\theta$ is the angle which the tangent at any point $t$ makes with the $X$ axis

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

$\therefore \theta =\frac{\pi }{4}+\frac{t}{2}=\frac{1}{4}(\pi +2t)$