If $x+iy=\frac{1+7i}{{\left(2-i\right)}^2};$ then $cosec\left({\tan }^{-1}\frac{y}{x}-\frac{\pi }{4}\right)=$

If $x+iy=\frac{1+7i}{{\left(2-i\right)}^2};$ then $cosec\left({\tan }^{-1}\frac{y}{x}-\frac{\pi }{4}\right)=$

Given, 

$x+iy=\frac{1+7i}{{\left(2-i\right)}^2}$

$x+iy=\frac{1+7i}{4+i^2-4i}$

$x+iy=\frac{1+7i}{4-1-4i}$

$x+iy=\frac{1+7i}{3-4i}$

Multiplying and dividing by $3-4i$

$x+iy=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}$

$x+iy=\frac{-25+25i}{25}$

$x+iy=-1+i$

Comparing on both sides

$x=-1, y=1$

$cosec\left({\tan }^{-1}\frac{y}{x}-\frac{\pi }{4}\right)$

$=cosec\left\{{\tan }^{-1}\left(-1\right)-\frac{\pi }{4}\right\}$

$=cosec\left(-\frac{\pi }{4}-\frac{\pi }{4}\right)$

$=cosec\left(-\frac{\pi }{2}\right)$

$=-1$