If $A=\int\limits^{sin \theta }_{1}\frac{t}{1+t^2}dt $ and $B = \int\limits^{cosec\theta }_{1}\frac{1}{t(1+t^2)}dt $, then the value of the determinant

$\begin{vmatrix}A & A^2 & B\\e^{A+B} & B^2 & -1\\1 & A^2+B^2 & -1\end{vmatrix}$ is

$0$

The correct answer is option (3) : $0$

We have,

$A+B=\int\limits^{sin \theta }_{1}\frac{t}{1+t^2}dt +\int\limits^{cosec\theta }_{t}\frac{1}{t(1+t^2)}dt $

$⇒A+B=\int\limits^{sin \theta }_{1}\frac{t}{1+t^2}dt + \int\limits^{sin \theta }_{1}-\frac{u}{1+u^2}, $ where $u=\frac{1}{t}$

$⇒ A + B= 0$

$⇒ B = - A$

$∴\begin{vmatrix}A & A^2 & B\\e^{A+B} & B^2 & -1\\1 & A^2+B^2 & -1\end{vmatrix}$

$=\begin{vmatrix}A & A^2 & -A\\1 & A^2 & -1\\1 & 2A^2 & -1\end{vmatrix}=-\begin{vmatrix}A & A^2 & A\\1 & A^2 & 1\\1 & 2A^2 & -1\end{vmatrix}=0$