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\(Let $A,$ $B,$ $C$ be the angles of a triangle. Evaluate \[\begin{vmatrix} \sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1 \end{vmatrix}.\]\)

 Apr 29, 2023
 #1
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We can use the Laplace expansion to evaluate the determinant of the given matrix. Expanding along the third column, we get:

| [[sin^2 A, cot A], [sin^2 B, cot B]] sin^2 C - [[sin^2 A, cot A], [sin^2 C, cot C]] sin^2 B + [[sin^2 B, cot B], [sin^2 C, cot C]] sin^2 A

where |M| denotes the determinant of matrix M. Note that the first term is the determinant of the 2x2 matrix [[sin^2 A, cot A], [sin^2 B, cot B]], which we can evaluate using the formula for the determinant of a 2x2 matrix:

[[sin^2 A, cot A], [sin^2 B, cot B]] = sin^2 A cot B - sin^2 B cot A

For the second and third terms, we can use the formula for the determinant of a 3x3 matrix:

[[a, b, c], [d, e, f], [g, h, i]] = a(ei - fh) - b(di - fg) + c(dh - eg)

Substituting the appropriate values, we get:

[[sin^2 A, cot A], [sin^2 C, cot C]] = sin^2 A cot C - sin^2 C cot A

[[sin^2 B, cot B], [sin^2 C, cot C]] = sin^2 B cot C - sin^2 C cot B

Substituting these values into our Laplace expansion, we get:

| [[sin^2 A, cot A], [sin^2 B, cot B], [sin^2 C, cot C]] = (sin^2 A cot B - sin^2 B cot A) sin^2 C

(sin^2 A cot C - sin^2 C cot A) sin^2 B

(sin^2 B cot C - sin^2 C cot B) sin^2 A

Factoring out sin^2 A, sin^2 B, and sin^2 C, respectively, we get:

| [[sin^2 A, cot A], [sin^2 B, cot B], [sin^2 C, cot C]] = sin^2 A (cot B sin^2 C - cot C sin^2 B)

sin^2 B (cot A sin^2 C - cot C sin^2 A)

sin^2 C (cot A sin^2 B - cot B sin^2 A)

Using the identity cot x = cos x / sin x, we can simplify the coefficients of cot A, cot B, and cot C:

cot B sin^2 C - cot C sin^2 B = cos B sin C - cos C sin B = sin(B - C) cot A sin^2 C - cot C sin^2 A = cos A sin C - cos C sin A = sin(A - C) cot A sin^2 B - cot B sin^2 A = cos A sin B - cos B sin A = sin(A - B)

Substituting these values, we get:

| [[sin^2 A, cot A], [sin^2 B, cot B], [sin^2 C, cot C]] = sin^2 A sin(B - C) - sin^2 B sin(A - C) + sin^2 C sin(A - B = 1.

Therefore, the determinant simplifies to 1.

 Apr 29, 2023
 #2
avatar+123 
+1

The question is  in LaTeX but the simplified version is :

 

 

Let A, B and C be the angles of a triangle. Evaluate:

 

 

                                                                                           \(\begin{vmatrix} \sin^2 A & \cot A & 1 \\ \sin^2 B & \cot B & 1 \\ \sin^2 C & \cot C & 1 \end{vmatrix}.\)

 May 1, 2023

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