If $\tan y = \frac{2t}{1 - t^2}$ and $\sin x = \frac{2t}{1 + t^2},$ then $\frac{dy}{dx} = $

Find $\frac{dy}{dx}$,if $x = a \cos \theta$ and $y = a \sin \theta$.

The slope of the tangent to the curve $x=t^{2}+3t-8$,$y=2t^{2}-2t-5$ at the point $(2,-1)$ is

If $x=e^\theta(\sin \theta-\cos \theta)$ and $y=e^\theta(\sin \theta+\cos \theta)$,then $\frac{dy}{dx}$ at $\theta=\frac{\pi}{4}$ is:

If $x = a(\cos \theta + \theta \sin \theta)$, $y = f(\theta)$, $f(2\pi) = 0$, $\frac{dy}{dx} = \frac{\tan \theta}{\theta}$, $\theta \neq 0$ and $\theta \neq (2n+1)\frac{\pi}{2}$, then $f\left(\frac{\pi}{3}\right) = $

If $x=\cos ^3 \theta-\sin ^3 \theta$ and $y=\sqrt[3]{\cos \theta}-\sqrt[3]{\sin \theta}$, then the value of $\frac{d y}{d x}$ at $\theta=\frac{\pi}{4}$ is