If A -1sin- t dt -1- t 2 dt - B -1cosec-1- t 1- t 2 dt thenA - A - 2 - B 1e - A - B - B - 2 -11 - A - 2- B - 2 -1lend vmatrix - A- sin-B- cosec-C- 0D- 1

The correct option is C 0B = ∫^cosec θ_1 1/t(1 + t^2) let 1/t = u ⇒ B = ∫_1^sin θ udu/1+u^2 ⇒ A + B = 0 ⇒ A = B ∴ A amp; A^2 amp; A e^0 amp;A^ ...

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Byju's Answer

Standard IX

Mathematics

Odd Extension of a Function

If A =∫1sinθ ...

Question

If $A = \int_{1}^{s i n θ} \frac{t d t}{1 + t^{2}} d t, B = \int_{1}^{c o s e c θ} \frac{1}{t (1 + t^{2})} d t$ then
$∣ ∣ ∣ ∣ ∣ \begin{matrix} A & A^{2} & B e^{A + B} & B^{2} & - 1 1 & A^{2} + B^{2} & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = ?$

s i n θ

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c o s e c θ

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0

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Solution

The correct option is C $0$
$B = \int_{1}^{c o s e c θ} \frac{1}{t (1 + t^{2})} l e t \frac{1}{t} = u \Rightarrow B = \int_{1}^{s i n θ} \frac{- u d u}{1 + u^{2}} \Rightarrow A + B = 0 \Rightarrow A = - B ∴ ∣ ∣ ∣ ∣ ∣ \begin{matrix} A & A^{2} & - A e^{0} & A^{2} & - 1 1 & A^{2} + B^{2} & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

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Similar questions

Q. If

A = \int_{1}^{s i n θ} \frac{t}{1 + t^{2}} d t

and

B = \int_{1}^{c o s e c θ} \frac{1}{t (1 + t^{2})} d t

, then the value of determinant

∣ ∣ ∣ ∣ ∣ \begin{matrix} A & A^{2} & B e^{A + B} & B^{2} & - 1 1 & A^{2} + B^{2} & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

Q. If

A = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣, B = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a^{2} & b^{2} & c^{2} a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣, C = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c a^{2} & b^{2} & c^{2} a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣,

then which relation is correct

$Δ_{1}$ , $∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{2} & b^{2} & c^{2} \end{matrix} ∣ ∣ ∣ ∣$ , $Δ_{2}$ = $∣ ∣ ∣ ∣ \begin{matrix} 1 & b c & a 1 & c a & b 1 & a b & c \end{matrix} ∣ ∣ ∣ ∣$ then

Q. If

∣ ∣ ∣ ∣ ∣ \begin{matrix} (a - 1)^{2} & a^{2} & (a + 1)^{2} (b - 1)^{2} & b^{2} & (b + 1)^{2} (c - 1)^{2} & c^{2} & (c + 1)^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣ = k ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & a & a^{2} 1 & b & b^{2} 1 & c & c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣,

then the value of

k

Q.
If

A = ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a & b & c a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣, B = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 a^{2} & b^{2} & c^{2} a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣, C = ∣ ∣ ∣ ∣ ∣ \begin{matrix} a & b & c a^{2} & b^{2} & c^{2} a^{3} & b^{3} & c^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣,

then which relation is correct

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If A=∫sinθ1t dt1+t2dt,B=∫cosecθ11t(1+t2)dt then ∣∣ ∣ ∣∣AA2BeA+BB2−11A2+B2−1∣∣ ∣ ∣∣=?

The correct option is C 0B=∫cosecθ11t(1+t2)let1t=u⇒B=∫sinθ1−udu1+u2⇒A+B=0⇒A=−B∴∣∣ ∣ ∣∣AA2−Ae0A2−11A2+B2−1∣∣ ∣ ∣∣=0

If $A = \int_{1}^{s i n θ} \frac{t d t}{1 + t^{2}} d t, B = \int_{1}^{c o s e c θ} \frac{1}{t (1 + t^{2})} d t$ then
$∣ ∣ ∣ ∣ ∣ \begin{matrix} A & A^{2} & B e^{A + B} & B^{2} & - 1 1 & A^{2} + B^{2} & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = ?$