Question

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

• A. 1
• B. $\infty$
• C. $-1$
• D. 0

Answer 1

Answer: (A)

$ =\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 $
$ \therefore \text{cosed}\left(\tan ^{-1} \frac{y}{x}-\frac{\pi}{4}\right)=\text{cosec}\left[\tan ^{-1}(-1)-\frac{\pi}{4}\right] $
$=\text{cosec}\left(\frac{3 \pi}{4}-\frac{\pi}{4}\right)$
$=\text{cosec} \frac{\pi}{2}=1$