Question

If $A=\int\limits_1^{\sin \theta} \frac{t}{1+t^2} d t, B=\int\limits_1^{\operatorname{cosec} \theta} \frac{1}{t\left(1+t^2\right)} d t$, then $\begin{vmatrix}A & A^2 & B \\ e^{A+B} & B^2 & -1 \\ 1 & A^2+B^2 & -1\end{vmatrix}$ equals

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

Answer 1

Answer: (C)

$B=\int\limits_1^{\operatorname{cosec} \theta} \frac{1}{t\left(1+t^2\right)} d t$
Let $\frac{1}{t}=u \Rightarrow B=\int\limits_1^{\sin \theta} \frac{-u}{1+u^2} d u $
$\Rightarrow A+B=0 \Rightarrow A=-B$
$\therefore\begin{vmatrix} A & A ^2 & B \\ e ^{ A + B } & B ^2 & -1 \\ 1 & A ^2+ B ^2 & -1\end{vmatrix}=0$