Q.The determinant $\begin{vmatrix} \sin A & \cos A & \sin A + \cos B \\ \sin B & \cos A & \sin B + \cos B \\ \sin C & \cos A & \sin C + \cos B \end{vmatrix}$ is equal to zero.

Split the third column: $C_3 = C_1 + (\cos B,\cos B,\cos B)^T$. Subtract $C_1$ from $C_3$ (this does not change the determinant):

$\begin{vmatrix} \sin A & \cos A & \sin A+\cos B \\ \sin B & \cos A & \sin B+\cos B \\ \sin C & \cos A & \sin C+\cos B \end{vmatrix} \xrightarrow{\,C_3\to C_3-C_1\,} \begin{vmatrix} \sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B \end{vmatrix}.$ …

The idea

A determinant is zero whenever two columns are proportional (linearly dependent). The awkward third column here is built to hide exactly that.

Simplify the third column

Write each third-column entry as its first-column part plus a common $\cos B$:

Apply the operation $C_3 \to C_3 - C_1$ (which leaves the determinant unchanged):

$\Delta = \begin{vmatrix} \sin A & \cos A & \cos B \\ \sin B & \cos A & \cos B \\ \sin C & \cos A & \cos B \end{vmatrix}.$ …

Method: Spotting a Zero Determinant by Exposing Proportional Columns

Before expanding a "clunky" determinant term by term, check whether a row/column operation reveals two proportional (or identical) rows or columns — that alone forces the determinant to $0$.

Steps

Step 1: Identify a row or column that is secretly a sum

Scan the entries for one that looks like "(something already in another column) + (a constant multiple of the all-ones vector)" or similar disguised repetition.

Step 2: Apply a row/column operation to strip away the disguise

Using operation 3 ($R_i\to R_i+\lambda R_j$ or the column analogue, $i\ne j$) never changes the determinant's value, so use it to subtract off the "already-present" part:

$\det \to \det \text{ with that column replaced by } (\text{original} - \text{matching column}).$ …

Common Mistakes

Mistake 1: Expanding the full $3\times3$ determinant term by term instead of spotting the structure

Why it's wrong: this determinant is deliberately built to hide a proportional-column shortcut; brute-force expansion is slow and much more likely to introduce an arithmetic slip. Correct approach: always check first whether a row/column operation exposes two proportional columns before committing to full expansion.

Mistake 2: Subtracting the wrong column when trying to simplify …