Question

Find the value of $x$ such that $\frac{(x+\alpha {)}^2-(x+\beta {)}^2}{\alpha +\beta }=\frac{sin2\theta }{si{n}^2\theta }$. when $\alpha$ and $\beta$ are the roots of $t^2-2t+2=0$

• A. $x=icot\theta -1$
• B. $x=-(icot\theta +1$)
• C. $x=icot\theta$
• D. $x=itan\theta -1$

Answer 1

The correct option is B $x=-(icot\theta +1$)

$t^2-2t+2=0$
$(t-1{)}^2-1+2=0$
$(t-1{)}^2=-1$
$t=1\pm i$
Hence
$\alpha +\beta =2$
$\alpha -\beta =2i$
Now let $n=2$
We get
$\frac{2x(\alpha -\beta )+{\alpha }^2-{\beta }^2}{2}=\frac{sin2\theta }{si{n}^2\theta }$
Hence
$\frac{4ix+2i-(-2i)}{2}=\frac{2sin\theta .cos\theta }{si{n}^2\theta }$
$2ix+2i=2cot\theta$
$ix=cot\theta -i$
$-x=icot\theta +1$
$x=-(1+icot\theta )$