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Matrices & Determinants: Value Determinant Vmatrix Vmatrix

JEE Maths question with a full step-by-step solution.

Question
The value of the determinant
∣2tan⁡Acot⁡B+cot⁡Atan⁡Btan⁡Acot⁡C+cot⁡Atan⁡Ctan⁡Bcot⁡A+cot⁡Btan⁡A2tan⁡Bcot⁡C+cot⁡Btan⁡Ctan⁡Ccot⁡A+cot⁡Ctan⁡Atan⁡Ccot⁡B+cot⁡Ctan⁡B2∣\begin{vmatrix} 2 & \tan A\cot B + \cot A\tan B & \tan A\cot C + \cot A\tan C \\ \tan B\cot A + \cot B\tan A & 2 & \tan B\cot C + \cot B\tan C \\ \tan C\cot A + \cot C\tan A & \tan C\cot B + \cot C\tan B & 2 \end{vmatrix}
 is:
A11
B22
C00correct
DNone of these
Solution
Step 1: Recognise the product structure Each off-diagonal entry (i,j)(i,j) has the form tan⁡Aicot⁡Aj+cot⁡Aitan⁡Aj\tan A_i\cot A_j + \cot A_i\tan A_j. The matrix can be expressed as:
M=(tan⁡Atan⁡Btan⁡C)(cot⁡Acot⁡Bcot⁡C)+(cot⁡Acot⁡Bcot⁡C)(tan⁡Atan⁡Btan⁡C)M = \begin{pmatrix}\tan A\\\tan B\\\tan C\end{pmatrix}\begin{pmatrix}\cot A&\cot B&\cot C\end{pmatrix} + \begin{pmatrix}\cot A\\\cot B\\\cot C\end{pmatrix}\begin{pmatrix}\tan A&\tan B&\tan C\end{pmatrix}
 Step 2: Express the determinant as a product
det⁡(M)=∣tan⁡Acot⁡A0tan⁡Bcot⁡B0tan⁡Ccot⁡C0∣×∣cot⁡Atan⁡A0cot⁡Btan⁡B0cot⁡Ctan⁡C0∣\det(M) = \begin{vmatrix}\tan A&\cot A&0\\\tan B&\cot B&0\\\tan C&\cot C&0\end{vmatrix} \times \begin{vmatrix}\cot A&\tan A&0\\\cot B&\tan B&0\\\cot C&\tan C&0\end{vmatrix}
 Both determinants contain a column of zeros, hence each equals zero.
det⁡(M)=0×0=0\det(M) = 0 \times 0 = 0
 Answer: (3)
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