If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x = a \cos \theta$ and $y = b \cos \theta$.

At any two points of the curve represented parametrically by $x = a(2 \cos t - \cos 2t)$ and $y = a(2 \sin t - \sin 2t)$,the tangents are parallel to the $x$-axis. The values of the parameter $t$ corresponding to these points differ from each other by:

$x=\cos \theta, y=\sin 5 \theta \Rightarrow (1-x^2) \frac{d^2 y}{d x^2}-x \frac{d y}{d x}$ is equal to (in $y$)

If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then:

If $x=\log _e\left(\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}}\right)$ and $\tan \frac{y}{2}=\sqrt{\frac{1-t}{1+t}}$,then the value of $\left(\frac{dy}{dx}\right)_{t=\frac{1}{2}}$ is

If $x = 2\cos t - \cos 2t$ and $y = 2\sin t - \sin 2t$,then at $t = \frac{\pi}{4}$,find $\frac{dy}{dx}$.