Question

If a curve is represented parametrically by the equations
$x=\sin (t+\frac{7\pi }{12})+\sin (t-\frac{\pi }{12})+\sin (t+\frac{3\pi }{12})$
$y=\cos (t+\frac{7\pi }{12})+\cos (t-\frac{\pi }{12})+\cos (t+\frac{3\pi }{12}),$
then the value of $\frac{d}{dt}(\frac{x}{y}-\frac{y}{x})$ at $t=\frac{\pi }{8}$ is

Answer 1

$x=\sin (t+\frac{7\pi }{12})+\sin (t-\frac{\pi }{12})+\sin (t+\frac{3\pi }{12})$
$=2\sin (t+\frac{\pi }{4})\cos \left(\frac{\pi }{3}\right)+\sin (t+\frac{\pi }{4})=\sin (t+\frac{\pi }{4})(2\cos \frac{\pi }{3}+1)=2\sin (t+\frac{\pi }{4})$

$y=\cos (t+\frac{7\pi }{12})+\cos (t-\frac{\pi }{12})+\cos (t+\frac{3\pi }{12})=2\cos (t+\frac{\pi }{4})\cos \left(\frac{\pi }{3}\right)+\cos (t+\frac{\pi }{4})=\cos (t+\frac{\pi }{4})(2\cos \frac{\pi }{3}+1)=2\cos (t+\frac{\pi }{4})$

$\frac{x}{y}=\tan (t+\frac{\pi }{4})=\frac{1+\tan t}{1-\tan t}$
and $\frac{y}{x}=\frac{1-\tan t}{1+\tan t}$

$\therefore (\frac{x}{y}-\frac{y}{x})=\left(\frac{1+\tan t}{1-\tan t}\right)-\left(\frac{1-\tan t}{1+\tan t}\right)=\frac{(1+\tan t{)}^2-(1-\tan t{)}^2}{1-{\tan }^{2}t}=\frac{4\tan t}{1-{\tan }^{2}t}=2\tan 2t$

Hence, $\frac{d}{dt}(\frac{x}{y}-\frac{y}{x})=\frac{d}{dt}\left(2\tan 2t\right)$
$={4{\sec }^{2}2t|}_{t=\pi /8}=4{\sec }^2\frac{\pi }{4}=8$