Question:
Using properties of determinants,prove that:
\(\begin{vmatrix}sinα& cosα& cos(α+δ)\\ sinβ& cosβ& cos(β+δ)\\ sinγ& cosγ& cos(γ+δ)\end{vmatrix}=0\)
Using properties of determinants,prove that:
\(\begin{vmatrix}sinα& cosα& cos(α+δ)\\ sinβ& cosβ& cos(β+δ)\\ sinγ& cosγ& cos(γ+δ)\end{vmatrix}=0\)
\(\begin{vmatrix}sinα& cosα& cos(α+δ)\\ sinβ& cosβ& cos(β+δ)\\ sinγ& cosγ& cos(γ+δ)\end{vmatrix}=0\)
Updated On: Sep 21, 2023
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Solution and Explanation
\(Δ\)\(\begin{vmatrix}sinα& cosα& cos(α+δ)\\ sinβ& cosβ& cos(β+δ)\\ sinγ& cosγ& cos(γ+δ)\end{vmatrix}\)
\(=\frac{1}{sinδcosδ}\begin{vmatrix}sinαsinδ& cosαcosδ& cosαcosδ-sinαsinδ\\ sinβsinδ& cosβcosδ& cosβcosδ-sinβsinδ\\ sinγsinδ& cosγcosδ& cosγcosδ-sinγsinδ\end{vmatrix}\)
Applying \(C_1→C_1+C_3\),we have
\(Δ=\frac{1}{sinδcosδ}\begin{vmatrix}cosαcosδ& cosαcosδ& cosαcosδ-sinαsinδ\\ cosβcosδ& cosβcosδ& cosβcosδ-sinβsinδ\\ cosγcosδ& cosγcosδ& cosγcosδ-sinγsinδ\end{vmatrix}\)
Here two columns \(C_1\) and \(C_2\) are identical
\( ∴Δ=0\)
Hence,the given result is proved.
\(=\frac{1}{sinδcosδ}\begin{vmatrix}sinαsinδ& cosαcosδ& cosαcosδ-sinαsinδ\\ sinβsinδ& cosβcosδ& cosβcosδ-sinβsinδ\\ sinγsinδ& cosγcosδ& cosγcosδ-sinγsinδ\end{vmatrix}\)
Applying \(C_1→C_1+C_3\),we have
\(Δ=\frac{1}{sinδcosδ}\begin{vmatrix}cosαcosδ& cosαcosδ& cosαcosδ-sinαsinδ\\ cosβcosδ& cosβcosδ& cosβcosδ-sinβsinδ\\ cosγcosδ& cosγcosδ& cosγcosδ-sinγsinδ\end{vmatrix}\)
Here two columns \(C_1\) and \(C_2\) are identical
\( ∴Δ=0\)
Hence,the given result is proved.
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