In triangle ABC- the value of e -i2A e iC e iB e iC e -i2B e iA e iB e iA e -i2C where i-1- is equal to

The correct option is C 4 Consider the given determinant. #160;D=| e ^ i2A amp; e ^ iC amp; e ^ iB e ^ iC amp; e ^ i2B amp; e ...

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Standard XII

Mathematics

Complex Numbers

In triangle ...

Question

In triangle $A B C$ , the value of $∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{- i 2 A} & e^{i C} & e^{i B} e^{i C} & e^{- i 2 B} & e^{i A} e^{i B} & e^{i A} & e^{- i 2 C} \end{matrix} ∣ ∣ ∣ ∣ ∣$ where $i = \sqrt{- 1}$ , is equal to

2

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4

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3

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none of these

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Solution

The correct option is C $4$
Consider the given determinant.
$D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{- i 2 A} & e^{i C} & e^{i B} e^{i C} & e^{- i 2 B} & e^{i A} e^{i B} & e^{i A} & e^{- i 2 C} \end{matrix} ∣ ∣ ∣ ∣ ∣$

Since, $A + B + C = π$
And $e^{i π} = cos π + i sin π = - 1$
$e^{i (B + C)} = e^{i (π - A)} = - e^{- i A} e^{- i (B + C)} = - e^{i A}$

On taking $e^{- i A}, e^{- i B}, e^{- i C}$ common from $R_{1}, R_{2}, R_{3}$ respectively, we get

$D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{- i A} & e^{i (A + C)} & e^{i (A + B)} e^{i (B + C)} & e^{- i B} & e^{i (A + B)} e^{i (B + C)} & e^{i (A + C)} & e^{- i C} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$D = ∣ ∣ ∣ ∣ ∣ \begin{matrix} e^{- i A} & {- e}^{- i B} & {- e}^{- i C} {- e}^{- i A} & e^{- i B} & {- e}^{- i C} {- e}^{- i A} & {- e}^{- i B} & e^{- i C} \end{matrix} ∣ ∣ ∣ ∣ ∣$
$D = e^{- i A} e^{- i B} e^{- i C} ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & - 1 - 1 & 1 & - 1 - 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ D = e^{- i (A + B + C)} ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & - 1 - 1 & 1 & - 1 - 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣$
$D = e^{- i π} ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & - 1 - 1 & 1 & - 1 - 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ D = - 1 ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & - 1 - 1 & 1 & - 1 - 1 & - 1 & 1 \end{matrix} ∣ ∣ ∣ ∣$
$D = - 1 \times [[1 (1 - 1) + 1 (- 1 - 1) - 1 (1 + 1)]] D = - 1 \times [0 - 2 - 2] D = - 1 \times - 4 D = 4$

Hence, the value of determinant is 4.

Therefore, option b is the correct option.

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Which of the following is the only correct combination?

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In triangle ABC, the value of ∣∣ ∣ ∣∣e−i2AeiCeiBeiCe−i2BeiAeiBeiAe−i2C∣∣ ∣ ∣∣ where i=√−1, is equal to

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