Without expanding at any stage, evaluate the value of determinant

$∣ ∣ ∣ ∣ \begin{matrix} 2 & t a n A c o t B + c o t A t a n B & t a n A c o t C + c o t A t a n C t a n B c o t A + c o t B t a n A & 2 & t a n B c o t C + c o t B t a n C t a n C c o t A + c o t C t a n A & t a n C c o t B + c o t C t a n B & 2 \end{matrix} ∣ ∣ ∣ ∣$

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Solution

$Δ = ∣ ∣ ∣ ∣ \begin{matrix} 2 & t a n A c o t B + c o t A t a n B & t a n A c o t C + c o t A t a n C t a n B c o t A + c o t B t a n A & 2 & t a n B c o t C + c o t B t a n C t a n C c o t A + c o t C t a n A & t a n C c o t B + c o t C t a n B & 2 \end{matrix} ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ \begin{matrix} t a n A c o t A + c o t A t a n A & t a n A c o t B + c o t A t a n B & t a n A c o t C + c o t A t a n C t a n B c o t A + c o t B t a n A & t a n B c o t B + c o t B t a n B & t a n B c o t C + c o t B t a n C t a n C c o t A + c o t C t a n A & t a n C c o t B + c o t C t a n B & t a n C c o t C + c o t C t a n C \end{matrix} ∣ ∣ ∣ ∣$

$= ∣ ∣ ∣ ∣ \begin{matrix} tan A & cot A & 0 tan B & cot B & 0 tan C & cot C & 0 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} cot A & tan A & 0 cot B & tan B & 0 cot C & tan C & 0 \end{matrix} ∣ ∣ ∣ ∣$

$= 0$

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