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Mathematics Question on Determinants

The value of the determinant ∣1cos;(α−β)cos;α cos(α−β)1cos;β cosαcos;β1∣\begin{vmatrix}1&cos; \left(\alpha-\beta\right)&cos; \alpha\\\ cos \left(\alpha -\beta \right)&1&cos; \beta\\\ cos \alpha &cos; \beta &1\end{vmatrix} is

A

α2+β2\alpha^{2}+\beta^{2}

B

α2−β2\alpha^{2}-\beta^{2}

C

11

D

00

Answer

00

Explanation

Solution

On solving the determinant,
1(1−cos⁡2β)−cos⁡(α−β)[cos⁡(α−β)−cos⁡α⋅cos⁡β]1\left(1-\cos ^{2} \beta\right)-\cos (\alpha-\beta)[\cos (\alpha-\beta)-\cos \alpha \cdot \cos \beta]
+cos⁡α[cos⁡β⋅cos⁡(α−β)−cos⁡α]+\cos \alpha[\cos \beta \cdot \cos (\alpha-\beta)-\cos \alpha]
=1−cos⁡2β−cos⁡2α−cos⁡2(α−β)= 1-\cos ^{2} \beta-\cos ^{2} \alpha-\cos ^{2}(\alpha-\beta)
+2cos⁡α⋅cos⁡β⋅cos⁡(α−β)+2 \cos \alpha \cdot \cos \beta \cdot \cos (\alpha-\beta)
=1−cos⁡2β−cos⁡2α+cos⁡(α−β)= 1-\cos ^{2} \beta-\cos ^{2} \alpha+\cos (\alpha-\beta)
[2cos⁡α⋅cos⁡β−cos⁡(α−β)][2 \cos \alpha \cdot \cos \beta-\cos (\alpha-\beta)]
=1−cos⁡2β−cos⁡2α+cos⁡(α−β)= 1-\cos ^{2} \beta-\cos ^{2} \alpha+\cos (\alpha-\beta)
[cos⁡(α+β)+cos⁡(α−β)−cos⁡(α−β)][\cos (\alpha+\beta)+\cos (\alpha-\beta)-\cos (\alpha-\beta)]
=1−cos⁡2β−cos⁡2α+cos⁡(α−β)cos⁡(α+β)= 1-\cos ^{2} \beta-\cos ^{2} \alpha+\cos (\alpha-\beta) \cos (\alpha+\beta)
=1−cos⁡2β−cos⁡2α+cos⁡2α⋅cos⁡2β−sin⁡2α⋅sin⁡2β= 1-\cos ^{2} \beta-\cos ^{2} \alpha+\cos ^{2} \alpha \cdot \cos ^{2} \beta-\sin ^{2} \alpha \cdot \sin ^{2} \beta
=1−cos⁡2β−cos⁡2α(1−cos⁡2β)−sin⁡2α⋅sin⁡2β= 1-\cos ^{2} \beta-\cos ^{2} \alpha\left(1-\cos ^{2} \beta\right)-\sin ^{2} \alpha \cdot \sin ^{2} \beta
=1−cos⁡2β−cos⁡2α⋅sin⁡2β−sin⁡2α⋅sin⁡2β= 1-\cos ^{2} \beta-\cos ^{2} \alpha \cdot \sin ^{2} \beta-\sin ^{2} \alpha \cdot \sin ^{2} \beta
=(1−cos⁡2β)−sin⁡2β(sin⁡2α+cos⁡2α)=\left(1-\cos ^{2} \beta\right)-\sin ^{2} \beta\left(\sin ^{2} \alpha+\cos ^{2} \alpha\right)
=sin⁡2β−sin⁡2β⋅1= \sin ^{2} \beta-\sin ^{2} \beta \cdot 1
=0= 0