Question

If $2\cos \alpha =x+\frac{1}{x},2\cos \beta =y+\frac{1}{y}$ then $\frac{x^{10}}{y^{12}}-\frac{y^{12}}{x^{10}}=$

• A. $\pm 2i\sin (10\alpha +12\beta )$
• B. $\pm i\sin (10\alpha +12\beta )$
• C. $\pm 2i\sin (10\alpha -12\beta )$
• D. $\pm i\sin (10\alpha -12\beta )$

Answer 1

Answer: (A), (C)

The correct options are
A $\pm 2i\sin (10\alpha +12\beta )$
C $\pm 2i\sin (10\alpha -12\beta )$

$2\cos \alpha =x+\frac{1}{x}$
$\Rightarrow x^2-2\left(\cos \alpha \right)x+1=0$
$\Rightarrow x=\frac{2\cos \alpha \pm \sqrt{4{\cos }^2\alpha -4}}{2}$
$\Rightarrow x=\cos \alpha \pm i\sin \alpha$
$\Rightarrow x=e^{\pm i\alpha }$
Similarly $y=e^{\pm i\beta }$
$\frac{x^{10}}{y^{12}}-\frac{y^{12}}{x^{10}}$
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Case $1:$ when $x=e^{i\alpha }$, then
$\frac{x^{10}}{y^{12}}-\frac{y^{12}}{x^{10}}=\frac{e^{10i\alpha }}{e^{\pm 12i\beta }}-\frac{e^{\pm 12i\beta }}{e^{10i\alpha }}=e^{i(10\alpha \mp 12\beta )}-e^{-i(10\alpha \mp 12\beta )}=2i\sin (10\alpha \mp 12\beta )=2i\sin (10\alpha +12\beta ),2i\sin (10\alpha -12\beta )$

Case $2:$ when $x=e^{-i\alpha }$, then
$\frac{x^{10}}{y^{12}}-\frac{y^{12}}{x^{10}}=\frac{e^{-10i\alpha }}{e^{\pm 12i\beta }}-\frac{e^{\pm 12i\beta }}{e^{-10i\alpha }}=e^{i(-10\alpha \mp 12\beta )}-e^{-i(-10\alpha \mp 12\beta )}=2i\sin (-10\alpha \mp 12\beta )=2i\sin (-10\alpha \mp 12\beta )=-2i\sin (10\alpha +12\beta ),-2i\sin (10\alpha -12\beta )$