If a -cos- i sin- b -cos- i sin-c -cos y - i sin Y and a - b - b - c - c - a -1- thenA- cos -cos -y-cos -0B- sin -sin - y -sin Y -0C- cos -cos - y -cos - -1D- sin -sin - y -sin v -1

Given, a=cosα+isinα=e^iα b=cosβ+isinβ=e^iβ c=cosγ+isinγ=e^iγ and ab+bc+ca=1 ⇒ e^i(α β)+e^i(β γ)+e^i(γ α)=1 ⇒cos(α β)+cos (β γ)+cos (γ α) +i[sin(α β)+sin (β γ) ...

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Standard XII

Mathematics

Properties of Argument

If a =cosα+ i...

Question

If $a = cos α + i sin α, b = cos β + i sin β,$
$c = cos γ + i sin γ$ and $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 1,$ then

cos (α - β) + cos (β - γ) + cos (γ - α) = 0

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sin (α - β) + sin (β - γ) + sin (γ - α) = 0

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cos (α - β) + cos (β - γ) + cos (γ - α) = 1

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sin (α - β) + sin (β - γ) + sin (γ - α) = 1

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Solution

The correct options are
B $sin (α - β) + sin (β - γ) + sin (γ - α) = 0$
C $cos (α - β) + cos (β - γ) + cos (γ - α) = 1$
Given,
$a = cos α + i sin α = e^{i α} b = cos β + i sin β = e^{i β}$
$c = cos γ + i sin γ = e^{i γ}$

and
$\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 1 \Rightarrow e^{i (α - β)} + e^{i (β - γ)} + e^{i (γ - α)} = 1 \Rightarrow cos (α - β) + cos (β - γ) + cos (γ - α) + i [sin (α - β) + sin (β - γ) + sin (γ - α)] = 1$

Equating real and imaginary parts of both the sides, we get
$\Rightarrow cos (α - β) + cos (β - γ) + cos (γ - α) = 1$
and
$sin (α - β) + sin (β - γ) + sin (γ - α) = 0$

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cos (α - β) + cos (β - γ) + cos (γ - α) = - \frac{3}{2}, cos α + cos β + cos γ = p & sin α + sin β + sin γ = q

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Properties of Argument

Standard XII Mathematics

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If a=cosα+isinα,b=cosβ+isinβ, c=cosγ+isinγ and ab+bc+ca=1, then

If $a = cos α + i sin α, b = cos β + i sin β,$
$c = cos γ + i sin γ$ and $\frac{a}{b} + \frac{b}{c} + \frac{c}{a} = 1,$ then