If $x = \sec \theta - \cos \theta$ and $y = \sec^n \theta - \cos^n \theta$,then $\left(\frac{dy}{dx}\right)^2$ is equal to

If $\theta$ is the angle made by the normal drawn to the curve $x=e^{t} \cos t, y=e^{t} \sin t$ at the point $(1,0)$,with the $X$-axis,then $\theta=$

If $x=a\left(t-\frac{1}{t}\right)$ and $y=b\left(t+\frac{1}{t}\right)$,then $\frac{dy}{dx}=$

Consider the curve represented parametrically by the equations $x = t^3 - 4t^2 - 3t$ and $y = 2t^2 + 3t - 5$,where $t \in \mathbb{R}$. If $H$ denotes the number of points on the curve where the tangent is horizontal and $V$ denotes the number of points where the tangent is vertical,then:

The second derivative of $a \sin^3 t$ with respect to $a \cos^3 t$ at $t = \frac{\pi}{4}$ is

If $x$ and $y$ are connected parametrically by the equations,without eliminating the parameter,find $\frac{dy}{dx}$ for $x=2at^{2}$ and $y=at^{4}$.