Question

If $\alpha ,\beta$ be the roots of the equation $u^2-2u+2=0$ and if $\cot \theta =x+1$, then $\frac{(x+\alpha {)}^n-(x+\beta {)}^n}{(\alpha -\beta )}$ is equal to

• A. $\frac{\sin n\theta }{{\sin }^n\theta }$
• B. $\frac{\cos n\theta }{{\cos }^n\theta }$
• C. $\frac{\sin n\theta }{{\cos }^n\theta }$
• D. $\frac{\cos n\theta }{{\sin }^n\theta }$

Answer 1

The correct option is A $\frac{\sin n\theta }{{\sin }^n\theta }$

$u^2-2u+2=0\Rightarrow u=1\pm i$
Let $\alpha =1+i,\beta =1-i$
$\Rightarrow \frac{(x+\alpha {)}^n-(x+\beta {)}^n}{\alpha -\beta }$
$=\frac{\left[\right(\cot \theta -1)+(1+i){]}^n-[(\cot \theta -1)+(1-i){]}^n}{2i}$
$(\because \cot \theta -1=x)$

$=\frac{(\cos \theta +i\sin \theta {)}^n-(\cos \theta -i\sin \theta {)}^n}{{\sin }^n\theta \left(2i\right)}$
$=\frac{2i\sin n\theta }{{\sin }^n\theta 2i}$
$=\frac{\sin n\theta }{{\sin }^n\theta }$