Q.Show that the $\triangle ABC$ is an isosceles triangle if the determinant $\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1 + \cos A & 1 + \cos B & 1 + \cos C \\ \cos^2 A + \cos A & \cos^2 B + \cos B & \cos^2 C + \cos C \end{vmatrix} = 0$.

Concept — reduce to a Vandermonde determinant. Row operations strip the determinant down to powers of $\cos A,\cos B,\cos C$; then $\Delta=0$ forces two cosines equal.

Start with

Apply $R_2\to R_2-R_1$ (middle row becomes $\cos A,\cos B,\cos C$), then $R_3\to R_3-R_2$ (last row becomes $\cos^2A,\cos^2B,\cos^2C$):

$\Delta=\begin{vmatrix}1&1&1\\\cos A&\cos B&\cos C\\\cos^2A&\cos^2B&\cos^2C\end{vmatrix}=(\cos B-\cos A)(\cos C-\cos A)(\cos C-\cos B).$ …

The idea

The rows are built from $1$, $1+\cos\theta$, and $\cos^2\theta+\cos\theta$. Subtracting neighbouring rows peels off the clutter and exposes the powers $1,\cos\theta,\cos^2\theta$ — the classic Vandermonde pattern, whose value is a product of differences.

Set up

Work the steps

1. $R_2\to R_2-R_1$. Each middle entry $1+\cos\theta$ loses its $1$:

2. $R_3\to R_3-R_2$. Each last entry $\cos^2\theta+\cos\theta$ loses its $\cos\theta$:

3. Vandermonde value. Since $\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix}=(y-x)(z-x)(z-y)$, with $x=\cos A,\,y=\cos B,\,z=\cos C$, $\Delta=(\cos B-\cos A)(\cos C-\cos A)(\cos C-\cos B).$ …

Method: Reducing to a Recognisable Standard Determinant (Vandermonde Pattern)

Use this method when a determinant's rows are built from repeated powers or shifted expressions of the same variables (here $1$, $1+\cos\theta$, $\cos^2\theta+\cos\theta$ for each angle) and the goal is to prove the determinant equals — or vanishes into — a simpler, recognisable expression.

Steps

Step 1: Strip each row down using row operations

Apply $R_i \to R_i - R_{i-1}$, working down the rows. Each subtraction removes the lower-degree part that was added onto that row's underlying variable, without changing the determinant's value (this is row operation 3 from the standard properties toolbox: adding a multiple of one row to a different row).

Step 2: Recognise the reduced form

After enough subtractions, the rows should reduce to a standard pattern — most commonly $1,\ x,\ x^2$ across each column for different values of $x$ (here $\cos A,\cos B,\cos C$). This is the classic Vandermonde determinant:

Step 3: Apply the standard factored value …

Common Mistakes

Mistake 1: Using the wrong row operations to reach the Vandermonde form

Why it's wrong: the reduction needs $R_2\to R_2-R_1$ followed by $R_3\to R_3-R_2$ (using the new $R_2$, not the original one) to correctly peel off the $1$ and then the $\cos\theta$ terms. Doing $R_3\to R_3-R_1$ instead leaves a $\cos A$ term tangled into row 3 and never produces the clean $\cos^2A,\cos^2B,\cos^2C$ row.

Mistake 2: Misremembering the Vandermonde sign/order

Why it's wrong: the identity $\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix}=(y-x)(z-x)(z-y)$ has a specific, non-symmetric factor order; swapping it to $(x-y)(y-z)(z-x)$ flips signs and can make a student misjudge which two angles end up equal when $\Delta=0$. …