Q.If $A + B + C = 0$, then prove that $\begin{vmatrix} 1 & \cos C & \cos B \\ \cos C & 1 & \cos A \\ \cos B & \cos A & 1 \end{vmatrix} = 0$

Key idea: expand the determinant, then use $A+B+C=0$ (so $\cos C=\cos(A+B)$) to collapse the trig.

Step 1 — expanding along the first row:

Step 2 — since $A+B+C=0$, $C=-(A+B)$ so $\cos C=\cos(A+B)$. Then …

Intuition

The determinant is a symmetric expression in $\cos A,\cos B,\cos C$. Expanding it produces a well-known combination, and the single condition $A+B+C=0$ is exactly what is needed to make that combination vanish, because it lets us write $\cos C=\cos(A+B)$.

Setting up

Working the steps

1. Expand along the first row:

Collecting terms,

2. Use the condition. From $A+B+C=0$ we get $C=-(A+B)$, so $\cos C=\cos(A+B)=\cos A\cos B-\sin A\sin B$. Group the terms containing $\cos C$:

$-\cos^2 C+2\cos A\cos B\cos C = \cos C\big(2\cos A\cos B-\cos C\big).$ …

Method: Direct Expansion + Trigonometric Substitution for Conditional Identities

Some determinant identities are proved not by row/column tricks but by expanding the determinant fully first, then using a given condition on the angles (like $A+B+C=0$ or $A+B+C=\pi$) to collapse the resulting trigonometric expression to a known value.

Steps

Step 1: Expand the determinant completely

For a symmetric $3\times3$ determinant whose entries are cosines (or a mix of $1$'s and cosines), expand along any row using the standard cofactor formula. The result is a fixed algebraic combination of the cosine terms — usually something of the form $1-\cos^2A-\cos^2B-\cos^2C+2\cos A\cos B\cos C$.

Step 2: Bring in the given angle condition

Use the stated relationship between the angles (e.g. $C=-(A+B)$ when $A+B+C=0$) to rewrite one of the cosines, typically $\cos C$, in terms of the other two using the compound-angle formula $\cos(A+B)=\cos A\cos B-\sin A\sin B$.

Step 3: Group and simplify using product-to-sum or Pythagorean identities …

Common Mistakes

Mistake 1: Dropping the minus sign in $\cos(A+B)$

Using $A+B+C=0$ gives $\cos C=\cos(A+B)=\cos A\cos B-\sin A\sin B$. Writing this with a $+$ instead of a $-$ breaks the later simplification to $\cos(A+B)\cos(A-B)$ and the whole proof stalls.

Mistake 2: Sign error in the cofactor expansion

Expanding the $3\times3$ determinant along the first row, the middle term carries a $-\cos C$ cofactor. Missing that sign changes the expanded form $1-\cos^2A-\cos^2B-\cos^2C+2\cos A\cos B\cos C$ into something that will never reduce to $0$. …