Without expanding, show that the value of each of the following determinants is zero :|(sin^2A,cotA,1)(sin^B,cotB,1)(sin^2C,cot C,1)|

Question

Without expanding, show that the value of each of the following determinants is zero :

\(\begin{vmatrix} sin^2A & cotA & 1 \\[0.3em] sin^2B & cotB &1 \\[0.3em] sin^2C & cot\,C & 1 \end{vmatrix}\), where A, B, C are the angles of ΔABC.

Answer 1

Let Δ = \(\begin{vmatrix} sin^2A & cotA & 1 \\[0.3em] sin^2B & cotB &1 \\[0.3em] sin^2C & cot\,C & 1 \end{vmatrix}\) 

Now, 

Δ = sin²A (cot B – cot C) – cot A (sin2B – sin²C) + 1 (sin²B cot C – cot B sin²C 

As A, B and C are angles of a triangle, 

A + B + C = 180°

Δ = sin²A cot B – sin²A cot C – cot A sin²B + cot A sin²C + sin²B cot C – cot B sin²C 

By using formulae,

Δ = 0 

Hence, Proved.