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If $\mathrm{a}=\cos 2 \alpha+\mathrm{i} \sin 2 \alpha, \mathrm{b}=\cos 2 \beta+\mathrm{i} \sin 2 \beta$, $c=\cos 2 \gamma+i \sin 2 \gamma$ and $d=\cos 2 \delta+i \sin 2 \delta$, then $\sqrt{\mathrm{abcd}}+\frac{1}{\sqrt{\mathrm{abcd}}}=$
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1919 Upvotes
Verified Answer
The correct answer is:
$2 \cos (\alpha+\beta+\gamma+\delta)$
We have,
$$
\begin{array}{l}
\begin{array}{l}
\text { abcd }=\cos (2 \alpha+2 \beta+2 \gamma+2 \delta)+i \sin (2 \alpha+2 \beta \\
+2 \gamma+2 \delta) \\
\therefore \sqrt{\text { abcd }}=[\cos (2 \alpha+2 \beta+2 \gamma+2 \delta) \\
+i \sin (2 \alpha+2 \beta+2 \gamma+2 \delta)]^{1 / 2} \\
\text { or } \sqrt{a b c d} \\
=\cos (\alpha+\beta+\gamma+\delta)+i \sin (\alpha+\beta+\gamma+\delta) \ldots . .(1)
\end{array}
\end{array}
$$
[De Moivre's Theorem]
$$
\begin{array}{l}
\therefore \frac{1}{\sqrt{\mathrm{abcd}}}=\cos (\alpha+\beta+\gamma+\delta)-\mathrm{i} \sin (\alpha+\beta \\
+\gamma+\delta).....(2)
\end{array}
$$
Adding (1) and (2), we obtain
$$
\sqrt{\mathrm{abcd}}+\frac{1}{\sqrt{\mathrm{abcd}}}=2 \cos (\alpha+\beta+\gamma+\delta)
$$
$$
\begin{array}{l}
\begin{array}{l}
\text { abcd }=\cos (2 \alpha+2 \beta+2 \gamma+2 \delta)+i \sin (2 \alpha+2 \beta \\
+2 \gamma+2 \delta) \\
\therefore \sqrt{\text { abcd }}=[\cos (2 \alpha+2 \beta+2 \gamma+2 \delta) \\
+i \sin (2 \alpha+2 \beta+2 \gamma+2 \delta)]^{1 / 2} \\
\text { or } \sqrt{a b c d} \\
=\cos (\alpha+\beta+\gamma+\delta)+i \sin (\alpha+\beta+\gamma+\delta) \ldots . .(1)
\end{array}
\end{array}
$$
[De Moivre's Theorem]
$$
\begin{array}{l}
\therefore \frac{1}{\sqrt{\mathrm{abcd}}}=\cos (\alpha+\beta+\gamma+\delta)-\mathrm{i} \sin (\alpha+\beta \\
+\gamma+\delta).....(2)
\end{array}
$$
Adding (1) and (2), we obtain
$$
\sqrt{\mathrm{abcd}}+\frac{1}{\sqrt{\mathrm{abcd}}}=2 \cos (\alpha+\beta+\gamma+\delta)
$$
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