Q.Value of the determinant |cos 67𝑜 sin 67𝑜 sin 23𝑜 cos 23𝑜| is
(A) 0
(B) 1 2
(C) √3 2
(D) 1

Concept: Determinant Evaluation Using Trigonometric Identities (complementary angles).

Step 1: Write the determinant:

Step 2: Use complementary angle relations: $\sin 23^\circ = \cos 67^\circ$ and $\cos 23^\circ = \sin 67^\circ$.

Step 3: Substitute: …

We need the value of

A $2\times 2$ determinant $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ equals $ad-bc$, so

This is exactly the cosine addition formula $\cos(A+B)=\cos A\cos B-\sin A\sin B$ with $A=67^\circ$, $B=23^\circ$:

$\Delta = \cos(67^\circ+23^\circ) = \cos 90^\circ = 0.$ …

Method: Complementary-Angle Symmetry to Collapse a Trigonometric Determinant

This method applies whenever a $2\times2$ (or larger) determinant is built from trigonometric ratios of two angles that are complementary (add to $90^\circ$) or otherwise related — the goal is to collapse the determinant using an identity rather than blind expansion.

Steps

Step 1: Expand the determinant using ad − bc

For $\begin{vmatrix}a&b\\c&d\end{vmatrix}$, always start by writing $ad-bc$ explicitly in terms of the given trig ratios. Do not evaluate individual trig values numerically yet — keep them symbolic so an identity can be spotted.

Step 2: Match the expansion to a standard trig identity

Once written as $\cos A\cos B-\sin A\sin B$ (or a similar pattern), recognise this as the addition/subtraction formula, e.g.

If the angles are complementary ($A+B=90^\circ$), the result collapses to $\cos 90^\circ=0$ immediately.

Step 3 (equivalent check): Use complementary-angle conversion to spot identical rows …

Common Mistakes

Mistake 1: Getting the complementary-angle identities backwards

Why it's wrong: students sometimes write $\sin 23^\circ=\sin 67^\circ$ or $\cos 23^\circ=\cos 67^\circ$ instead of the correct complementary relations $\sin(90^\circ-\theta)=\cos\theta$ and $\cos(90^\circ-\theta)=\sin\theta$, which breaks the row-matching that makes the determinant collapse to zero. Correct approach: since $23^\circ=90^\circ-67^\circ$, use $\sin 23^\circ=\cos 67^\circ$ and $\cos 23^\circ=\sin 67^\circ$ before touching the determinant.

Mistake 2: Slipping on the sign in the cosine addition formula

Why it's wrong: expanding $\cos 67^\circ\cos 23^\circ-\sin 67^\circ\sin 23^\circ$ directly, a student may recall $\cos(A-B)$ (with a $+$ sign) instead of $\cos(A+B)$ (with a $-$ sign), giving $\cos 44^\circ$ instead of $\cos 90^\circ$. Correct approach: the determinant expansion $ad-bc$ already carries the minus sign, so it matches $\cos(A+B)=\cos A\cos B-\sin A\sin B$ exactly — recognise this pattern rather than re-deriving it from scratch. …