If A=∫ limits1 sin θ (t/1+t2) d t, B=∫ limits1 operatornamecosec θ (1/t(1+t2)) d t, then |A A2 B eA+B B2 -1 1 A2+B2 -1| equals

Question

If A=∫ limits1 sin θ (t/1+t2) d t, B=∫ limits1 operatornamecosec θ (1/t(1+t2)) d t, then |A A2 B eA+B B2 -1 1 A2+B2 -1| equals (A)  sin θ (B)  operatornamecosec θ (C) 0 (D) 1

Tardigrade Admin · Accepted Answer

B=∫ limits1 operatornamecosec θ (1/t(1+t2)) d t Let (1/t)=u ⇒ B=∫ limits1 sin θ (-u/1+u2) d u ⇒ A+B=0 ⇒ A=-B ∴| A A 2 B e A + B B 2 -1 1 A 2+ B 2 -1|=0

Q. If $A = 1 \int s i n θ \frac{t}{1 + t ^{2}} d t, B = 1 \int cosec θ \frac{1}{t ( 1 + t ^{2} )} d t$ , then $∣ ∣ A e^{A + B} 1 A^{2} B^{2} A^{2} + B^{2} B - 1 - 1 ∣ ∣$ equals

$sin θ$

$cosec θ$

0

1

$B = 1 \int cosec θ \frac{1}{t ( 1 + t ^{2} )} d t$
Let $\frac{1}{t} = u \Rightarrow B = 1 \int s i n θ \frac{- u}{1 + u ^{2}} d u$
$\Rightarrow A + B = 0 \Rightarrow A = - B$
$∴ ∣ ∣ A e^{A + B} 1 A^{2} B^{2} A^{2} + B^{2} B - 1 - 1 ∣ ∣ = 0$

Q. If A=1∫sinθ​1+t2t​dt,B=1∫cosecθ​t(1+t2)1​dt, then ∣∣​AeA+B1​A2B2A2+B2​B−1−1​∣∣​ equals

sinθ

cosecθ

0

1

B=1∫cosecθ​t(1+t2)1​dt Let t1​=u⇒B=1∫sinθ​1+u2−u​du ⇒A+B=0⇒A=−B ∴∣∣​AeA+B1​A2B2A2+B2​B−1−1​∣∣​=0

Q. If $A = 1 \int s i n θ \frac{t}{1 + t ^{2}} d t, B = 1 \int cosec θ \frac{1}{t ( 1 + t ^{2} )} d t$ , then $∣ ∣ A e^{A + B} 1 A^{2} B^{2} A^{2} + B^{2} B - 1 - 1 ∣ ∣$ equals

$sin θ$

$cosec θ$

$B = 1 \int cosec θ \frac{1}{t ( 1 + t ^{2} )} d t$
Let $\frac{1}{t} = u \Rightarrow B = 1 \int s i n θ \frac{- u}{1 + u ^{2}} d u$
$\Rightarrow A + B = 0 \Rightarrow A = - B$
$∴ ∣ ∣ A e^{A + B} 1 A^{2} B^{2} A^{2} + B^{2} B - 1 - 1 ∣ ∣ = 0$