If a-cos- -isin- - b-cos- -isin- - c-cos- -isin-a-b-b-c-c-a-1- Then cos- - -cos- - -cos- - -

Explanation for correct option.Step1. Given that,a=cosα +isinα , ....(i)0ex0exb=cosβ +isinβ , ....(ii)0ex0exc=cosγ +isinγ , ....(iii)0ex0ex1= a/b+b/c+c/a .. ...

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Byju's Answer

Standard XII

Mathematics

Continuity of a Function

If a=cosα +is...

Question

If $a = \cos α + i \sin α, b = \cos β + i \sin β, c = \cos γ + i \sin γ & \frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 1$ . Then $\cos (α - β) + \cos (β - γ) + \cos (γ - α) = .$

$\frac{3}{2}$

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$0$

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$\frac{- 3}{2}$

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$1$

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Solution

The correct option is D
$1$

Explanation for correct option.
Step1. Given that,
$a = \cos α + i \sin α, . . . . (i) b = \cos β + i \sin β, . . . . (i i) c = \cos γ + i \sin γ, . . . . (i i i) 1 = \frac{a}{b} + \frac{b}{c} + \frac{c}{a} . . . . . (i v)$
Step2. Finding the value of $\cos (α - β) + \cos (β - γ) + \cos (γ - α)$
$a = e^{i α}, b = e^{i β}, c = e^{i γ}$
From equation (iv)
$e^{i (α - β)} + e^{i (β - γ)} + e^{i (γ - α)} = 1 \cos (α - β) + i \sin (α - β) + \cos (β - γ) + i \sin (β - γ) + \cos (γ - α) + i \sin (γ - α) = 1$
Arrange real & imaginary part together
$\cos (α - β) + \cos (β - γ) + \cos (γ - α) + i [\sin (α - β) + \sin (β - γ) + \sin (γ - α)] = 1 + 0 i$
Comparing real parts of both side of equal to sign, we get
$\cos (α - β) + \cos (β - γ) + \cos (γ - α) = 1$
Hence, correct option is $(D)$

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If a=cosα+isinα,b=cosβ+isinβ,c=cosγ+isinγ&ab+bc+ca=1. Then cos(α-β)+cos(β-γ)+cos(γ-α)=.

If $a = \cos α + i \sin α, b = \cos β + i \sin β, c = \cos γ + i \sin γ & \frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 1$ . Then $\cos (α - β) + \cos (β - γ) + \cos (γ - α) = .$