Let $A = \begin{bmatrix} \cos\alpha & -\sin\alpha & 0 \\ \sin\alpha & \cos\alpha & 0 \\ 0 & 0 & 1 \end{bmatrix}$. If $B = \text{adj}\,A$, then the matrix $B^{-1}$ is equal to...

Let $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 6 & -13 \\ 5 & -10 \end{bmatrix}$ be two matrices. If the variables $x$ and $y$ satisfy the matrix equation $((A^{-1})^2 + B)\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$, then the ordered pair $(x, y) = $

If $0 \leq x \leq 1$, $I_1 = \int\sin^{-1}\sqrt{1-x^2}\,dx$ and $I_2 = \int\sin^{-1}x\,dx$, then which of the following is true?

Let $t \in (0, 1)$ and $\alpha \in \left(0, \dfrac{\pi}{4}\right)$. If $x = \text{cosec}^{-1}\left(\dfrac{1+t^2}{2t}\right)$, $y = \cot^{-1}\left(\dfrac{\sqrt{1-t^2}}{t}\right)$ and $\dfrac{dy}{dx} = f(t)$, then the value of $f(\tan\alpha)$ is