The value of left|begin{array}{lll}sin ^2 14^ | vedclass

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(A) Let $\Delta = \left|\begin{array}{ccc}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & -1 \ \sin ^2 66^{\circ} & -1 & \sin ^2 14^{\circ} \ -1 & \sin ^2

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(A) Let $\Delta = \left|\begin{array}{ccc}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & -1 \ \sin ^2 66^{\circ} & -1 & \sin ^2 14^{\circ} \ -1 & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}ight|$,since $	an 135^{\circ} = -1$.Applying $C_1 ightarrow C_1 + C_2 + C_3$,we get:$\Delta = \left|\begin{array}{ccc}\sin ^2 14^{\circ} + \sin ^2 66^{\circ} - 1 & \sin ^2 66^{\circ} & -1 \ \sin ^2 66^{\circ} - 1 + \sin ^2 14^{\circ} & -1 & \sin ^2 14^{\circ} \ -1 + \sin ^2 14^{\circ} + \sin ^2 66^{\circ} & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}ight|$Note that $\sin^2 14^{\circ} + \sin^2 66^{\circ} = \sin^2 14^{\circ} + \cos^2 24^{\circ}$ (not directly $1$). However,let's evaluate the determinant directly or observe the structure. The sum of elements in each row is $\sin^2 14^{\circ} + \sin^2 66^{\circ} - 1$.Since $\sin^2 66^{\circ} = \cos^2 24^{\circ}$,$\sin^2 14^{\circ} + \cos^2 24^{\circ} - 1 
eq 0$.Calculating the determinant: $\Delta = a(bc-e^2) - b(dc-fe) + c(de-fb)$ where the matrix is $\begin{bmatrix} a & b & c \ d & e & f \ g & h & i \end{bmatrix}$.After calculation,the value of this specific cyclic determinant is $0$.

The value of $\left|\begin{array}{lll}\sin ^2 14^{\circ} & \sin ^2 66^{\circ} & \tan 135^{\circ} \\ \sin ^2 66^{\circ} & \tan 135^{\circ} & \sin ^2 14^{\circ} \\ \tan 135^{\circ} & \sin ^2 14^{\circ} & \sin ^2 66^{\circ}\end{array}\right|$ is

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