If \(ABC\) is a triangle, then what is the value of the determinant
\(\begin{vmatrix}\cos C & \sin B & 0 \\\tan A & 0 & \sin B \\0 & \tan(B+C) & \cos C\end{vmatrix}?\)
Triangle Determinant Calculation
The problem asks us to find the value of the determinant for a triangle \(ABC\). The determinant involves trigonometric functions of the angles \(A\), \(B\), and \(C\). We know that for any triangle, the sum of the angles is \(A + B + C = \pi\) radians (or \(180^\circ\)). This property will be key to solving the problem.
Evaluating the Determinant
The determinant given is:
\(D = \begin{vmatrix} \cos C & \sin B & 0 \\ \tan A & 0 & \sin B \\ 0 & \tan(B+C) & \cos C \end{vmatrix}\)
Step 1: Expand the Determinant
We can calculate the value of this 3x3 determinant by expanding it. Let's use the cofactor expansion along the first row:
\(D = \cos C \begin{vmatrix} 0 & \sin B \\ \tan(B+C) & \cos C \end{vmatrix} - \sin B \begin{vmatrix} \tan A & \sin B \\ 0 & \cos C \end{vmatrix} + 0 \begin{vmatrix} \tan A & 0 \\ 0 & \tan(B+C) \end{vmatrix}\)
Step 2: Calculate the 2x2 Determinants
Now, we compute the values of the two 2x2 determinants:
- First 2x2 determinant:
\(\begin{vmatrix} 0 & \sin B \\ \tan(B+C) & \cos C \end{vmatrix} = (0 \times \cos C) - (\sin B \times \tan(B+C)) = -\sin B \tan(B+C)\)
- Second 2x2 determinant:
\(\begin{vmatrix} \tan A & \sin B \\ 0 & \cos C \end{vmatrix} = (\tan A \times \cos C) - (\sin B \times 0) = \tan A \cos C\)
Step 3: Substitute and Simplify
Substitute the results from Step 2 back into the expansion from Step 1:
\(D = \cos C (-\sin B \tan(B+C)) - \sin B (\tan A \cos C) + 0\)
Simplifying this expression gives:
\(D = -\cos C \sin B \tan(B+C) - \sin B \tan A \cos C\)
Step 4: Apply Triangle Angle Property
For any triangle \(ABC\), we know that the sum of the angles is \(\pi\) radians:
\(A + B + C = \pi\)
From this, we can express \(B+C\) in terms of \(A\):
\(B + C = \pi - A\)
Now, let's find the tangent of \((B+C)\):
\(\tan(B+C) = \tan(\pi - A)\)
Using the trigonometric identity \(\tan(\pi - \theta) = -\tan \theta\), we get:
\(\tan(B+C) = -\tan A\)
Step 5: Final Determinant Value
Substitute the result \(\tan(B+C) = -\tan A\) into the simplified expression for \(D\) from Step 3:
\(D = -\cos C \sin B (-\tan A) - \sin B \tan A \cos C\)
\(D = \cos C \sin B \tan A - \sin B \tan A \cos C\)
The two terms cancel each other out:
\(D = 0\)
Result
Therefore, the value of the given determinant for a triangle \(ABC\) is 0.
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