Question

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

• A. $\sin \theta$
• B. $cosec\theta$
• C. $0$
• D. $1$

Answer 1

The correct option is C $0$

$A={\int }_1^{sin\theta }\frac{t}{1+t^2}dt$
$B={\int }_1^{cosec\theta }\frac{1}{t(1+t^2)}dt$
Put $z=\frac{1}{t}$
$dz=-\frac{1}{t^2}dt$
$B={\int }_1^{\sin \theta }\frac{-z}{(z^2+1)}dz$
$={\int }_1^{\sin \theta }\frac{-t}{(t^2+1)}dt$
$\Rightarrow B=-A$
Now, $\left|\begin{array}{ccc}A& A^2& B\\ e^{A+B}& B^2& -1\\ 1& A^2+B^2& -1\end{array}\right|$
$=\left|\begin{array}{ccc}A& A^2& -A\\ e^0& A^2& -1\\ 1& 2A^2& -1\end{array}\right|$
$=-\left|\begin{array}{ccc}A& A^2& A\\ 1& A^2& 1\\ 1& 2A^2& 1\end{array}\right|$
$=0$ (On expanding the determinant)