Question

Using properties of determinants,prove that:
$\begin{vmatrix}sinα& cosα& cos(α+δ)\\\ sinβ& cosβ& cos(β+δ)\\\ sinγ& cosγ& cos(γ+δ)\end{vmatrix}=0$

Answer 1

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.