If x+i y=(1+7 i/(2-i)2) ; then operatornamecosec( operatornameTan-1 (y/x)-(π/4))=

Question

If x+i y=(1+7 i/(2-i)2) ; then operatornamecosec( operatornameTan-1 (y/x)-(π/4))= (A) 1 (B) ∞ (C) -1 (D) 0

Tardigrade Admin · Accepted Answer

=(1+7 i/3-4 i) × (3+4 i/3+4 i) =(25(-1+i)/25) x+i y=-1+i ⇒ x=-1, y=1 ∴ textcosed( tan -1 (y/x)-(π/4))= textcosec[ tan -1(-1)-(π/4)] = textcosec((3 π/4)-(π/4)) = textcosec (π/2)=1

Q. If $x + i y = \frac{1 + 7 i}{( 2 - i ) ^{2}};$ then $cosec (Tan^{- 1} \frac{y}{x} - \frac{π}{4}) =$

1

$\infty$

$- 1$

0

$= \frac{1 + 7 i}{3 - 4 i} \times \frac{3 + 4 i}{3 + 4 i}$
$= \frac{25 ( - 1 + i )}{25}$
$x + i y = - 1 + i$
$\Rightarrow x = - 1, y = 1$
$∴ cosed (tan^{- 1} \frac{y}{x} - \frac{π}{4}) = cosec [tan^{- 1} (- 1) - \frac{π}{4}]$
$= cosec (\frac{3 π}{4} - \frac{π}{4})$
$= cosec \frac{π}{2} = 1$

Q. If x+iy=(2−i)21+7i​; then cosec(Tan−1xy​−4π​)=

1

∞

−1

0

=3−4i1+7i​×3+4i3+4i​ =2525(−1+i)​ x+iy=−1+i ⇒x=−1,y=1 ∴cosed(tan−1xy​−4π​)=cosec[tan−1(−1)−4π​] =cosec(43π​−4π​) =cosec2π​=1

Q. If $x + i y = \frac{1 + 7 i}{( 2 - i ) ^{2}};$ then $cosec (Tan^{- 1} \frac{y}{x} - \frac{π}{4}) =$

$\infty$

$- 1$

$= \frac{1 + 7 i}{3 - 4 i} \times \frac{3 + 4 i}{3 + 4 i}$
$= \frac{25 ( - 1 + i )}{25}$
$x + i y = - 1 + i$
$\Rightarrow x = - 1, y = 1$
$∴ cosed (tan^{- 1} \frac{y}{x} - \frac{π}{4}) = cosec [tan^{- 1} (- 1) - \frac{π}{4}]$
$= cosec (\frac{3 π}{4} - \frac{π}{4})$
$= cosec \frac{π}{2} = 1$