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

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

If the tangent to the curve given by $x=t^{2}-1$ and $y=t^{2}-t$ is parallel to the $X$-axis,then the value of $t$ is

If $x=a(\cos t+t \sin t)$ and $y=a(\sin t-t \cos t),$ find $\frac{d^{2} y}{d x^{2}}.$

The derivative of $f(\sec x)$ with respect to $g(\tan x)$ at $x=\frac{\pi}{4}$,where $f^{\prime}(\sqrt{2})=4$ and $g^{\prime}(1)=2$,is

If $x=\sec \theta-\cos \theta$,$y=\sec ^{10} \theta-\cos ^{10} \theta$ and $(x^2+4)(\frac{dy}{dx})^2=k(y^2+4)$,then the value of $k$ is