fA, B, C are the angles of a triangle, then

Key Concepts

Linear Dependence in Matrices

A key concept in linear algebra is that if the rows or columns of a matrix are linearly dependent, the determinant of the matrix is zero. In problems involving angles of a triangle and their trigonometric functions, inherent relationships (such as those arising from the triangle angle sum property) may cause such dependencies, thereby simplifying the computation of the determinant.

Trigonometric Functions and Identities

Trigonometric functions such as cosine and cotangent, and their squared and reciprocal forms, are essential in relating the angles of a triangle to various geometric and algebraic expressions. The use of these functions in the matrix elements connects the geometry of the triangle with algebraic properties like determinant evaluation.

Triangle Angle Sum Property

In Euclidean geometry, the sum of the angles in any triangle is always ? radians (180 degrees). This fundamental property is often used to establish relationships among trigonometric functions of the angles in a triangle, which can then influence the behavior of expressions involving these functions.

Matrix Determinant

The determinant of a matrix is a scalar value that can be computed from its elements and has important properties, such as indicating whether the rows or columns of the matrix are linearly independent. In the context of the problem, the determinant being zero implies that the rows (or columns) of the matrix are linearly dependent.

Transcript

00:01 So in the given question we are told that a, b and c are angles of a triangle, right? they are angles of a triangle and if that's the case we are told to evaluate the determinant cause square a cost square b cost square c cot a cot b cot c one one one so this is the determinant we need to evaluate right so what we can do over here is let's take the determinant and do some row operations over here right so we can change r2 to r2 minus r1 and r3 to r3 minus r3 minus r2 right? so first let's change r3 minus r2 right so r3 changes to r3 minus r2 would give us the determinant as for square a, cot a 1, cause square b, cot b, cot b, 1 cost square c minus b cost square c minus cost square b cross square b cross square c minus o square b cot c minus cot b and 1 minus 1 which is equal to 0 in the next step what we can do is we can do the row operation r2 changes to r2 minus r1 and then we would have the determinant cost square a, cost square b minus cost square a, cost square c minus cost square b.

02:14 And over here we have cot a, cot b minus cot a and cot c minus cot b.

02:26 And in the last row we have 0 1 0 0 so this is what we have now we can just expand this right but what we should know is two identities right so we should know the identity that when we are taking cos square x minus cos square y we can change we can write cos square x minus cost square y we can change we can write cos square x minus cost square y as sine square y minus sine square x right so we can write it as sine square y minus sine square x and there is a relation that says the sign of x plus y times the sign of y minus x is equal to sine square y minus sine square x so we are going to to use this relation over here right and next what we can do is we have to know another property which is the cot of x minus the cot of y can be evaluated as the sign of y minus x divided by the sign of x times the sign of y so these are two properties that we need to know right so we are going to use this properties to simplify the simply simplify the determinant further so then we can write cos square b minus cost square a in the matrix let's write the first row which is unchanged cos square a and for cost square b minus cost square a we can write sign of a plus b times sign of a minus b.

04:39 For cos square c minus cause square b in the third row, we can write sign of b plus c times sign of b minus c.

04:55 And then we have the second column where it is cot a.

05:01 And for cot b minus cot a, we can write it as.

05:06 Sine of a minus b divided by sine a, sine a times sine b.

05:16 And the third element in this row, in this column is cot b minus cot c, which would then be equal to sine of b minus c divided by sine b times sine c the rest of the elements are the same that is one zero zero and now what we can do is we can evaluate this determinant right so what we would have is sign of a plus b sine of a plus b times sine of a minus b multiplied by sign of sine of b minus c divided by sign of b minus c minus minus we have sign of b plus c times b minus c times sine of a minus b divided by by sine of a times sine of b.

06:46 So this is what we get when we expand the determinant, right? this is what we get when we expand the determinant...

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