In $\triangle ABC$, with usual notations, if $\Delta$ denotes the area of triangle $ABC$ then the value of $2s(b+c-a)\tan\left(\dfrac{A}{2}\right)$ is equal to $\ldots$

If matrix A and its inverse $A^{-1}$ are given by $A = \begin{bmatrix} 0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & x & 1 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} \dfrac{1}{2} & -\dfrac{1}{2} & \dfrac{1}{2} \\ -4 & 3 & y \\ \dfrac{5}{2} & -\dfrac{3}{2} & \dfrac{1}{2} \end{bmatrix}$, then the polar co-ordinates of the points whose Cartesian co-ordinates are $(x, y)$ are $\ldots$

Let $A = \begin{bmatrix} -3 & 2 \\ 1 & 4 \end{bmatrix}$ and if $A^2 - 2A + I = \begin{bmatrix} 18 & p \\ q & 11 \end{bmatrix}$, then $\ldots$

For $x > 0$ and $(x \log x) < 1$, if $y = \cot^{-1}\left(\dfrac{x - \log x^{x^2}}{\log e^{x^2} + \log x^x}\right)$, then $\dfrac{dy}{dx} = \ldots$