What is value of \(\begin{vmatrix}sin^2 ⁡a & sina\, cosa & cos^2 ⁡a \\sin^2 ⁡b & sinb\, cosb & cos^2 ⁡b \\sin^2⁡ c & sinc\, cosc & cos^2⁡ c \end {vmatrix}\)?

We have, \(\begin{vmatrix}sin^2 a & sina \,cosa & cos^2 a\\sin^2 b & sinb\, cosb & cos^2 b\\sin^2 c & sinc\, cosc & cos^2 c \end {vmatrix}\)
Now, multiplying by 2 in both numerator and denominator of column 2 and C₁ = C₁ + C₃ we get,
1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & 2sina\, cosa & cos^2 a\\sin^2 b + cos^2 b & 2sinb\, cosb & cos^2 b\\sin^2 c + cos^2 c & 2sinc\, cosc & cos^2 c \end {vmatrix}\)
= 1/2 \(\begin{vmatrix}sin^2 a + cos^2 a & sin2a & cos^2 a\\sin^2 b + cos^2 b & sin2b & cos^2 b\\sin^2 c + cos^2 c & sin2c & cos^2 c \end {vmatrix}\)
= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2 a\\1 & sin2b & cos^2 b\\1 & sin2c & cos^2 c \end {vmatrix}\)
Solving further,
= 1/2 \(\begin{vmatrix}1 & sin2a & cos^2?a \\0 & sin2b-sin2a & cos^2? b-cos^2 ?a \\0 & sin2c-sin2a & cos^2 ?c-cos^2 ?a \end {vmatrix}\)
= 1/2 [(sin2b – sin2a)(cos²?c – cos²?a) ? (cos² b – cos²a)(sin2c – sin2a)]
Now, since, [cos² ?A + cos² B = sin(A + B) * sin(B ? A)]
So, 1/2 [2 cos(a + b) sin(b ? a) * sin(c + a)sin(a ? c) ? sin(a + b)sin(a ? b) * 2 cos(a + c)sin(c ? a)]
= sin(a ? b)sin(c ? a)[sin(c + a)cos(a + b) ? cos(c + a) sin(a + b)]
= sin(a ? b) sin(c ? a) sin(c + a ? a ? b)
= -sin(a ? b) sin(b ? c) sin(c ? a)