JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Hard

1 of 1

Application +4 | -1

a=\cos 2 \alpha+i \sin 2 \alpha, \quad b=\cos 2 \beta+i \sin 2 \beta

c=\cos 2 \gamma+i \sin 2 \gamma

and

d=\cos 2\delta+i \sin 2\delta

, then

\sqrt{a b c d}+\frac{1}{\sqrt{a b c d}}

is equal to

View answer and explanation

Correct option: B

2 \cos (\alpha+\beta+\gamma+\delta)

Option B (

2 \cos (\alpha+\beta+\gamma+\delta)

) is correct because by applying De Moivre's theorem and simplifying, the expression equals

2 \cos (\alpha+\beta+\gamma+\delta)

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