$\text{ If }A={\int }_1^{\sin \theta }\frac{t}{1+t^2}dt,B={\int }_1^{\cos e\theta }\frac{1}{t\left(1+t^2\right)}dt,\text{ then the value of }\left|\begin{array}{ccc}A& A^2& B\\ e^{A+B}& B^2& -1\\ 1& A^2+B^2& -1\end{array}\right|$is

$B=\underset{1}{\overset{\cos ec\theta }{\int }}\frac{1}{t\left(1+t^2\right)}dt$

$\text{ Put }t=\frac{1}{y}\Rightarrow dt=-\frac{dy}{y^2}$

$\text{ So, }B={\int }_1^{\sin \theta }\frac{1}{\frac{1}{y}\left(1+\frac{1}{y^2}\right)}\left(-\frac{dy}{y^2}\right)=-{\int }_1^{\sin \theta }\frac{y}{1+y^2}dy$

$\text{ We get }A+B=0$

$\Rightarrow \left|\begin{array}{ccc}A& A^2& B\\ e^{A+B}& B^2& -1\\ 1& A^2+B^2& -1\end{array}\right|=\left|\begin{array}{ccc}A& A^2& -A\\ 1& A^2& -1\\ 1& 2A^2& -1\end{array}\right|=0$