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Question: If A = \(\int_{1}^{\sin\theta}\frac{tdt}{1 + t^{2}}\) and B = \(\int_{1}^{\cos ec\theta}\frac{dt}{t(...

If A = ∫1sin⁡θtdt1+t2\int_{1}^{\sin\theta}\frac{tdt}{1 + t^{2}} and B = ∫1cos⁡ecθdtt(1+t2)\int_{1}^{\cos ec\theta}\frac{dt}{t(1 + t^{2})} then the value of ∣AA2BeAeBB2−11A2+B2−1∣\left| \begin{matrix} A & A^{2} & B \\ e^{A}e^{B} & B^{2} & - 1 \\ 1 & A^{2} + B^{2} & - 1 \end{matrix} \right| is

A

sinq

B

cosecq

C

0

D

1

Answer

0

Explanation

Solution

A = ∫1sin⁡θtdt1+t2\int_{1}^{\sin\theta}\frac{tdt}{1 + t^{2}}

B = ∫1cos⁡ecθdtt(1+t2)\int_{1}^{\cos ec\theta}\frac{dt}{t(1 + t^{2})} Put t = 1z\frac{1}{z}, dt = – 1z2\frac{1}{z^{2}} dz

B = ∫1sin⁡θ−1z2dz1z(1+1z2)\int_{1}^{\sin\theta}\frac{- \frac{1}{z^{2}}dz}{\frac{1}{z}\left( 1 + \frac{1}{z^{2}} \right)}

B = ∫1sin⁡θ−z(z2+1)\int_{1}^{\sin\theta}\frac{- z}{(z^{2} + 1)}dz

B = – ∫1sin⁡θz(z2+1)\int_{1}^{\sin\theta}\frac{z}{(z^{2} + 1)}dz z Ž t

B = – ∫1sin⁡θtt2+1\int_{1}^{\sin\theta}\frac{t}{t^{2} + 1}dt = – A

B = – A A + B = 0

Now D = ∣AA2−A1A2−112A2−1∣\left| \begin{matrix} A & A^{2} & - A \\ 1 & A^{2} & - 1 \\ 1 & 2A^{2} & - 1 \end{matrix} \right| = 0