If $A =\int\limits_{1}^{\sin θ}\frac{t}{1+t^2}dt$ and $B=\int\limits_{1}^{cosec θ}\frac{1}{t(1+t^2)}dt$, 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
We have,
$A+B=\int\limits_{1}^{\sin θ}\frac{t}{1+t^2}dt+\int\limits_{1}^{cosec θ}\frac{1}{t(1+t^2)}dt$
$⇒A+B=\int\limits_{1}^{\sin θ}\frac{t}{1+t^2}dt+\int\limits_{1}^{\sin θ}-\frac{u}{1+u^2}du$, where $u=\frac{1}{t}$
$⇒A+B=0$
$⇒B=-A$
$∴\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}$
$=\begin{vmatrix}A&A^2&-A\\1&A^2&-1\\1&2A^2&-1\end{vmatrix}=-\begin{vmatrix}A&A^2&A\\1&A^2&1\\1&2A^2&1\end{vmatrix}=0$