Question

If $\alpha ,\beta$ are the roots of the equation $u^2-2u+2=0$ and if $\cot \theta =x+1$, then $\left[\right(x+\alpha {)}^n-(x+\beta {)}^m]/[\alpha -\beta ]$ 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 $\frac{\sin n\theta }{{\sin }^n\theta }$

$u^2-2u+2=0$
$⟹u=1\pm i$
So,$\alpha =1+iand\beta =1-i$
Now given that,$x=\cot \theta -1$
so,$\frac{{(x+\alpha )}^n-{(x+\beta )}^n}{\alpha -\beta }=\frac{{(\cot \theta -1+1+i)}^n-(\cot \theta -1+1-i{)}^n}{2i}$
$=\frac{{(\cot \theta +i)}^n-(\cot \theta -i{)}^n}{2i}=\frac{{(\cos \theta +i\sin \theta )}^n-{(\cos \theta -\sin \theta )}^n}{\left(\sin \theta {)}^n\right(2i)}$
$=\frac{e^{\left(in\theta \right)}-e^{-\left(in\theta \right)}}{({\sin \theta )}^{n}2i}$
=$\frac{(\cos \left(n\theta \right)+i\sin \left(n\theta \right))-(\cos \left(n\theta \right)-i\sin \left(n\theta \right))}{({\sin \theta )}^{n}2i}=\frac{\sin \left(n\theta \right)}{{(\sin \theta )}^n}$