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Sin(A+B)Sin(A-B)

cosθ+α sinθ+...

Question

$∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (θ + α) & sin (θ + α) & 1 cos (θ + β) & sin (θ + β) & 1 cos (θ + γ) & sin (θ + γ) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$ is independent of

A

α

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B

β

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C

γ

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D

θ

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Solution

The correct option is D $θ$
Differentiating the given determinant w.r.t. $θ$ we get
$∣ ∣ ∣ ∣ ∣ \begin{matrix} - sin (θ + α) & sin (θ + α) & 1 - sin (θ + β) & sin (θ + β) & 1 - sin (θ + γ) & sin (θ + γ) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ + ∣ ∣ ∣ ∣ ∣ \begin{matrix} cos (θ + α) & cos (θ + α) & 1 cos (θ + β) & cos (θ + β) & 1 cos (θ + γ) & cos (θ + γ) & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= 0 + 0 = 0$
where $f (θ)$ denotes the given determinant
$\Rightarrow f (θ)$ is independent of $θ$ .
Alternately expanding the determinant along last column we get
$sin (θ + γ - θ - β) - sin (θ + γ - θ - α) + sin (θ + β - θ - α)$
$= sin (γ - β) + sin (β - α) + sin (α - γ)$
which is independent of $θ$ .

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Similar questions

α + β + γ = 2 θ

cos θ + cos (θ - α) + cos (θ - β) + cos (θ - γ)

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cos (θ - α) = a

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, then prove that

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