Question

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

$x^2-2x+2=0\&\cot \theta =x+1$
$\alpha$ and $\beta$ are the roots of the equation.
$\Rightarrow \alpha ,\beta =1+i,1-i$
$x+\alpha =\cot \theta +i$
$x+\alpha =cosec\theta \left(cis\right(\theta \left)\right)$
$(x+\alpha {)}^n=(cosec\theta {)}^n\left(cis\right(n\theta \left)\right)$
Similarly,
$(x+\beta {)}^n=(cosec\theta {)}^n\left(cis\right(-n\theta \left)\right)$
Now,
$(x+\alpha {)}^n-(x+\beta {)}^n$
$=\left(cosec\theta {)}^n\right(cis\left(n\theta \right)-cis(-n\theta ))$
$=\left(cosec\theta {)}^n\right(2i\sin \left(n\theta \right))$
$=\frac{2i\sin \left(n\theta \right)}{(\sin \theta {)}^n}$

Also, $\alpha -\beta =2$.

Hence, Given expression is $\frac{2i\frac{\sin \left(n\theta \right)}{(\sin \theta {)}^n}}{2i}$, which is $=\frac{\sin \left(n\theta \right)}{(\sin \theta {)}^n}$
Hence, option A is correct.