VITEEE 2016 Paper

We have,
abcd = cos (2α + 2β + 2γ + 2δ) + i sin (2α + 3β + 2γ + 2δ)
∴ √abcd = [cos(2α+2β+2γ+2δ) + isin(2α+2β+2γ+2δ)]1∕2
or √abcd = cos (α + β + γ + δ) + i sin (α + 2β + γ + δ) ... (1)
[De Moivre's Theorem]
∴

1

√abcd

= cos (α + β + γ + δ) - i sin (α + β + γ + δ) ... (2)
Adding (1) and (2), we obtain
√abcd + 1/*√abcd = 2 cos (α + β + γ + δ)