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Mathematics Question on Properties of Determinants

The value of the determinant ∣sin⁡236∘cos⁡236∘cot⁡135∘ sin⁡253∘cot⁡135∘cos⁡253∘ cot⁡135∘cos⁡225∘cos⁡265∘∣\begin{vmatrix}\sin ^{2} 36^{\circ} & \cos ^{2} 36^{\circ} & \cot 135^{\circ} \\\ \sin ^{2} 53^{\circ} & \cot 135^{\circ} & \cos ^{2} 53^{\circ} \\\ \cot 135^{\circ} & \cos ^{2} 25^{\circ} & \cos ^{2} 65^{\circ}\end{vmatrix} is

A

-2

B

-1

C

0

D

1

Answer

0

Explanation

Solution

Δ=∣sin⁡236∘cos⁡236∘cot⁡135∘ sin⁡253∘cot⁡135∘cos⁡253∘ cot⁡135∘cos⁡225∘cos⁡265∘∣\Delta=\begin{vmatrix}\sin ^{2} 36^{\circ} & \cos ^{2} 36^{\circ} & \cot 135^{\circ} \\\ \sin ^{2} 53^{\circ} & \cot 135^{\circ} & \cos ^{2} 53^{\circ} \\\ \cot 135^{\circ} & \cos ^{2} 25^{\circ} & \cos ^{2} 65^{\circ}\end{vmatrix}
∴cos⁡236∘=cos⁡2(90−54)=sin⁡254∘\therefore \cos ^{2} 36^{\circ}=\cos ^{2}(90-54)=\sin ^{2} 54^{\circ}
sin⁡237∘=sin⁡2(90−53)∘=cos⁡253∘\sin ^{2} 37^{\circ}=\sin ^{2}(90-53)^{\circ}=\cos ^{2} 53^{\circ}
cos⁡265∘=cos⁡2(90−25)∘=sin⁡225∘\cos ^{2} 65^{\circ}=\cos ^{2}(90-25)^{\circ}=\sin ^{2} 25^{\circ}
cot⁡135∘=cot⁡(90+45)∘=−tan⁡45∘=−1\cot 135^{\circ}=\cot (90+45)^{\circ}=-\tan 45^{\circ}=-1
∴Δ=∣cos⁡254∘sin⁡254∘−1 sin⁡253−1cos⁡253∘ −1cos⁡225∘sin⁡225∘∣\therefore \Delta=\begin{vmatrix}\cos ^{2} 54^{\circ} & \sin ^{2} 54^{\circ} & -1 \\\ \sin ^{2} 53 & -1 & \cos ^{2} 53^{\circ} \\\ -1 & \cos ^{2} 25^{\circ} & \sin ^{2} 25^{\circ}\end{vmatrix}
Δ=∣cos⁡254∘+sin⁡254∘−1sin⁡254∘−1 sin⁡253∘−1+cos⁡253∘−1cos253∘ −1+cos⁡225∘+sin⁡225∘cos⁡225∘sin⁡225∘∣\Delta=\begin{vmatrix}\cos ^{2} 54^{\circ}+\sin ^{2} 54^{\circ}-1 \sin ^{2} 54^{\circ} -1\\\ \sin ^{2} 53^{\circ}-1+\cos ^{2} 53^{\circ}-1 cos^{2} 53^{\circ} \\\ -1+\cos ^{2} 25^{\circ}+\sin ^{2} 25^{\circ} \cos ^{2} 25^{\circ} \sin ^{2} 25^{\circ} \end{vmatrix}
=∣0sin⁡254∘−1 0−1cos⁡253∘ 0cos⁡225∘sin⁡225∘∣=0=\begin{vmatrix}0 & \sin ^{2} 54^{\circ} & -1 \\\ 0 & -1 & \cos ^{2} 53^{\circ} \\\ 0 & \cos ^{2} 25^{\circ} & \sin ^{2} 25^{\circ}\end{vmatrix}=0