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
Answer & explanation
Correct answer: option 3
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 $u=\frac{1}{t}$
$⇒ A + B= 0$
$⇒ B = - A$
$∴\begin{vmatrix}A & A^2 & B\\e^{A+B} & B^2 & -1\\1 & A^2+B^2 & -1\end{vmatrix}$
$=\begin{vmatrix}A & A^2 & -A\\1 & A^2 & -1\\1 & 2A^2 & -1\end{vmatrix}=-\begin{vmatrix}A & A^2 & A\\1 & A^2 & 1\\1 & 2A^2 & -1\end{vmatrix}=0$