Question

If $a=cos2\alpha +isin2\alpha ,b=cos2\beta ,c=cos2\gamma +isin2\gamma$ and $d=cos2\delta +isin2\delta$, then $\sqrt{abcd}+\frac{1}{\sqrt{abcd}}$ is equal to

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is B

$abcd=cos(2\alpha +2\beta +2\gamma +2\delta )+isin(2\alpha +2\beta +2\gamma +2\delta )\therefore \sqrt{abcd}={\left[cos2\right(\alpha +\beta +\gamma +\delta )+isin2(\alpha +\beta +\gamma +\delta \left)\right]}^{\frac{1}{2}}\dots \dots \left(i\right)$
[deMoivre's Theorem]
$\therefore \frac{1}{\sqrt{abcd}}=cos(\alpha +\beta +\gamma +\delta )+isincos(\alpha +\beta +\gamma +\delta )\dots \dots \left(ii\right)$
On adding Equations. (i) and (ii), we obtain
$\sqrt{abcd}+\frac{1}{\sqrt{abcd}}=2cos(\alpha +\beta +\gamma +\delta )$