Question

If $2\cos \alpha =a+\left(\frac{1}{a}\right)$ and $2\cos \beta =b+\left(\frac{1}{b}\right),$ then $ab+\left(\frac{1}{ab}\right)$ is equal to

• A. $2cos(\alpha +\beta )$
• B. $2\sin (\alpha +\beta )$
• C. $2cos(\alpha -\beta )$
• D. $4\cos \alpha \cos \beta$

Answer 1

Answer: (A)

Given, $2\cos \alpha =a+\frac{1}{a}$
$\Rightarrow$ $a^2-2a\cos \alpha +1=0$ ...(i)
and $2\cos \beta =b+\frac{1}{b}$
$\Rightarrow$ $b^2-2b\cos \beta +1=0$ ...(ii)
From Eq. (i),
$a=\frac{2\cos \alpha \pm \sqrt{4{\cos }^2\alpha -4}}{2}$
$\Rightarrow$ $a=\frac{2\cos \alpha \pm 2i\sin \alpha }{2}$
$\Rightarrow$ $a=\cos \alpha \pm i\sin \alpha$
Similarly, $b=cos\beta \pm isin\beta$
$\therefore$ $ab=(\cos \alpha \pm i\sin \alpha )(\cos \beta \pm i\sin B)$
$=\cos (\alpha +\beta )\pm i\sin (\alpha +\beta )$
and $\frac{1}{ab}=\cos (\alpha +\beta )\mp i\sin (\alpha +\beta )$
Hence, $ab+\frac{1}{ab}=2\cos (\alpha +\beta )$