Question

The value of the determinant $\begin{vmatrix}\sin ^{2} 36^{\circ} & \cos ^{2} 36^{\circ} & \cot 135^{\circ} \\ \sin ^{2} 53^{\circ} & \cot 135^{\circ} & \cos ^{2} 53^{\circ} \\ \cot 135^{\circ} & \cos ^{2} 25^{\circ} & \cos ^{2} 65^{\circ}\end{vmatrix}$ is

• A. -2
• B. -1
• C. 0
• D. 1

Answer 1

The Correct Option is C

$\Delta=\begin{vmatrix}\sin ^{2} 36^{\circ} & \cos ^{2} 36^{\circ} & \cot 135^{\circ} \\ \sin ^{2} 53^{\circ} & \cot 135^{\circ} & \cos ^{2} 53^{\circ} \\ \cot 135^{\circ} & \cos ^{2} 25^{\circ} & \cos ^{2} 65^{\circ}\end{vmatrix}$
$\therefore \cos ^{2} 36^{\circ}=\cos ^{2}(90-54)=\sin ^{2} 54^{\circ}$
$\sin ^{2} 37^{\circ}=\sin ^{2}(90-53)^{\circ}=\cos ^{2} 53^{\circ}$
$\cos ^{2} 65^{\circ}=\cos ^{2}(90-25)^{\circ}=\sin ^{2} 25^{\circ}$
$\cot 135^{\circ}=\cot (90+45)^{\circ}=-\tan 45^{\circ}=-1$
$\therefore \Delta=\begin{vmatrix}\cos ^{2} 54^{\circ} & \sin ^{2} 54^{\circ} & -1 \\ \sin ^{2} 53 & -1 & \cos ^{2} 53^{\circ} \\ -1 & \cos ^{2} 25^{\circ} & \sin ^{2} 25^{\circ}\end{vmatrix}$
$\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} $
$=\begin{vmatrix}0 & \sin ^{2} 54^{\circ} & -1 \\ 0 & -1 & \cos ^{2} 53^{\circ} \\ 0 & \cos ^{2} 25^{\circ} & \sin ^{2} 25^{\circ}\end{vmatrix}=0$