First slide

Introduction to Determinants

Question

If A, B and C are angles of triangle, then the value of $|\begin{array}{ccc} \sin^{2} A & \cot A & 1 \\ \sin^{2} B & \cot B & 1 \\ \sin^{2} C & \cot C & 1 \end{array}|$ is

Moderate

A

tanA + tanB + C
B

cotA cotB cotC
C

sin²A + sin²B + sin²C
D

0

Solution

Applying $R_{2} \to R_{2} - R_{1} and R_{3} \to R_{3} - R_{1}$ , we get
$Δ = |\begin{array}{ccc} \sin^{2} A & \cot A & 1 \\ \sin (B + A) \sin (B - A) & \frac{\sin (A - B)}{\sin Asin B} & 0 \\ \sin (C + A) \sin (C - A) & \frac{\sin (A - C)}{\sin Asin C} & 0 \end{array}|$ $[∵ \cot α - \cot β = \frac{\sin (β - α)}{\sin αsin β}]$
Expanding along C₃, we get
$\begin{matrix} Δ = \frac{\sin (A - B) \sin (A - C)}{\sin A} [- \frac{\sin (B + A)}{\sin C} + \frac{\sin (C + A)}{\sin B}] \\ = \frac{\sin (A - B) \sin (A - C)}{\sin A} [- \frac{\sin (π - C)}{\sin C} + \frac{\sin (π - B)}{\sin B}] \\ = \frac{\sin (A - B) \sin (A - C)}{\sin A} [- \frac{\sin C}{\sin C} + \frac{\sin B}{\sin B}] = 0 \end{matrix}$

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