If $x = \sqrt{2^{\csc^{-1} t}}$ and $y = \sqrt{2^{\sec^{-1} t}}$ for $|t| \ge 1$,then $\frac{dy}{dx}$ is equal to:

$x=\cos ^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right), y=\sin ^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \Rightarrow \frac{d y}{d x}$ is equal to

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:

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 $x = 4t^3 + 3$, $y = 3t^4 + 4$ and $\frac{d^2x}{dy^2} = (\frac{dx}{dy})^n$ is constant, then the value of $n$ is

For $a > 0, t \in \left( 0, \frac{\pi}{2} \right)$,let $x = \sqrt{a^{\sin^{-1} t}}$ and $y = \sqrt{a^{\cos^{-1} t}}$. Then,$1 + \left( \frac{dy}{dx} \right)^2$ equals