Question

The value of $\begin{vmatrix} \sin 30^{\circ} & \cos 30^{\circ} & \sin(30^{\circ}+75^{\circ}) \\ \sin 45^{\circ} & \cos 45^{\circ} & \sin(45^{\circ}+75^{\circ}) \\ \sin 60^{\circ} & \cos 60^{\circ} & \sin(60^{\circ}+75^{\circ}) \end{vmatrix}$ is equal to

• A. -2
• B. -1
• C. 0
• D. 1
• E. 2

Hint:

Logic Tip: In matrix problems involving trigonometric sum or difference formulas across rows or columns, look for linear dependence immediately rather than calculating and evaluating each sine/cosine value individually.

Answer 1

The Correct Option is C

Concept:
A determinant evaluates to zero if any column (or row) can be expressed as a linear combination of the other columns (or rows). The compound angle formula $\sin(A + B) = \sin A \cos B + \cos A \sin B$ is key to identifying this relationship.
Step 1: Expand the terms in the third column.
Let the third column be $C_3$. Applying the sum formula for sine, where $B = 75^{\circ}$: Row 1: $\sin(30^{\circ}+75^{\circ}) = \sin 30^{\circ} \cos 75^{\circ} + \cos 30^{\circ} \sin 75^{\circ}$ Row 2: $\sin(45^{\circ}+75^{\circ}) = \sin 45^{\circ} \cos 75^{\circ} + \cos 45^{\circ} \sin 75^{\circ}$ Row 3: $\sin(60^{\circ}+75^{\circ}) = \sin 60^{\circ} \cos 75^{\circ} + \cos 60^{\circ} \sin 75^{\circ}$
Step 2: Express $C_3$ as a linear combination of $C_1$ and $C_2$.
Observe the first two columns: $C_1$ consists of $\sin \theta$ values. $C_2$ consists of $\cos \theta$ values. From Step 1, every element in $C_3$ is formed exactly by: $$C_3 = (\cos 75^{\circ}) \cdot C_1 + (\sin 75^{\circ}) \cdot C_2$$
Step 3: Apply determinant properties.
Since the third column is merely a linear combination of the first two columns, the columns are linearly dependent. According to the properties of determinants, if one column is a linear combination of the others, the determinant is exactly equal to 0. Alternatively, perform the column operation $C_3 \to C_3 - (\cos 75^{\circ})C_1 - (\sin 75^{\circ})C_2$. This transforms the entire third column into zeros, yielding a determinant of 0.